Moving Charges And Magnetism Class 12th Physics Part I CBSE Solution

Class 12^th Physics Part I CBSE Solution

Exercises

1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current…
2. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a…
3. A long straight wire in the horizontal plane carries a current of 50 A in north to south…
4. A horizontal overhead power line carries a current of 90 A in east to west direction. What…
5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8…
6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its…
7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the…
8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The…
9. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil…
10. Two moving coil meters, M1 and M2 have the following particulars: R1 = 10Ω, N1 = 30, A1 =…
11. In a chamber, a uniform magnetic field of 6.5 G (1 G = 10-4 T) is maintained. An electron…
12. In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit.…
13. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended…
14. Would your answer change, if the circular coil in (a) were replaced by a planar coil of…

Additional Exercises

1. Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the…
2. A magnetic field of 100 G (1 G = 10-4 T) is required which is uniform in a region of…
3. For a circular coil of radius R and N turns carrying current I, the magnitude of the…
4. A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm,…
5. A magnetic field that varies in magnitude from point to point but has a constant direction…
6. A charged particle enters an environment of a strong and non-uniform magnetic field…
7. An electron travelling west to east enters a chamber having a uniform electrostatic field…
8. An electron emitted by a heated cathode and accelerated through a potential difference of…
9. A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a…
10. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two…
11. The wires which connect the battery of an automobile to its starting motor carry a current…
12. A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its…
13. A uniform magnetic field of 3000 G is established along the positive z-direction. A…
14. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10…
15. A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A…
16. A galvanometer coil has a resistance of 12 and the metre shows full scale deflection for a…
17. A galvanometer coil has a resistance of 15 and the metre shows full scale deflection for a…

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Exercises

Question 1.

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer:

Given:

Number of turns, n = 100

Radius of coil, r = 8 cm

Current through the coil, I = 0.40 A

Magnitude of magnetic field at centre of coil, B = ?

Using Biot Savart law, we find that magnetic field at the centre of a coil is given by,

 …(1)

Where,

B = Magnetic field strength

n = total number of turns

I = current through the coil

μ₀ is the permeability of free space.

μ₀ = 4 × π × 10^-7 TmA^-1

r = radius of coil

Now, by putting the values in equation (1), we get

⇒ |B| = 3.14 × 10^-4T

∴ Magnitude of magnetic field at the centre of the coil is 3.14 × 10⁴T.

Question 2.

A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Answer:

Given:

Current through the wire, I = 35 A

Distance of point P from the wire, d = 20 cm

Using Biot Savart law we can find magnetic field at a distance d from a current carrying conductor.

It is given by:

 …(1)

Where,

B = Magnetic field strength

I = current through the coil

�₀ is the permeability of free space.

�₀ = 4 × π × 10^-7 TmA^-1

d = distance of point P

By plugging the values in the equation (1), we get

⇒ 

⇒ |B| = 3.5 × 10^-5T

In this case the magnetic field at 20 cm away from a current carrying conductor is found to be 3.5 × 10^-5T.

Question 3.

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Answer:

Given:

Current through the wire, I = 50 A (North to South)

Distance of point P East of the wire, d = 2.5 m

Using Biot Savart law we can find magnetic field at a distance d from a current carrying conductor.

It is given by:

 …(1)

Where,

B = Magnetic field strength

I = current through the coil

�₀ is the permeability of free space.

�₀ = 4 × π × 10^-7 TmA^-1

d = distance of point P

By plugging the values in the equation (1), we get

⇒ |B| = 4 × 10^-6T

Direction of magnetic field,

The point is in a plane normal to the wire and the wire carries current in north to south. Using Right hand thumb rule we can conclude that the direction of magnetic field is vertically upwards, or out of the paper.

The magnitude of the magnetic field is 4 × 10^-6T and its direction is upwards or out of paper.

Question 4.

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer:

Given:

Current through the wire, I = 90 A (East to West)

Distance of point P below the wire, d = 1.5 m

Using Biot Savart law we can find magnetic field at a distance d from a current carrying conductor.

It is given by:

 …(1)

Where,

B = Magnetic field strength

I = current through the coil

�₀ is the permeability of free space.

�₀ = 4 × π × 10^-7 TmA^-1

d = distance of point P

By plugging the values in the equation (1), we get

⇒ 

⇒ |B| = 1.2 × 10^-5T

Direction of magnetic field,

We know that wire carries current in east to west direction. Using Right hand thumb rule, we can conclude that the direction of magnetic field is from north to south as indicated in the figure.

The magnitude of the magnetic field is 1.2 × 10^-5T and its direction is from north to south.

Question 5.

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Answer:

Given:

Current through the wire, I = 8A

Strength of magnetic field = 0.15T

Angle between direction of magnetic field and current, θ = 30°

Force on a current carrying conductor of length L carrying current I, due to a uniform magnetic field of strength B is given by,

F = IL × B …(1)

In the equation we assume length to be unity (L = 1) to calculate the force per unit length.

From equation (1) we have,

F = IB×sin(θ) …(2)

Now by putting the values in equation (2), we get

F = 8A × 0.15T × sin30°

⇒ F = 0.6 Nm^-1

Hence, the magnitude of force per unit length of the wire is 0.6 Nm^-1

Question 6.

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer:

Given:

Current through the wire, I = 10A

Strength of magnetic field inside the solenoid = 0.27T

Angle between direction of magnetic field and current, θ = 90°

Length of the wire = 3 cm

Force on a current carrying conductor of length L carrying current I, due to a uniform magnetic field of strength B is given by,

F = IL × B …(1)

Now by plugging the values in the equation (1), we get,

F = 10A × 0.03m × 0.27T × sin90°

⇒ F = 8.1 × 10^-2 N

Hence, the magnitude of force per unit length of the wire is 8.1 × 10^-2 N.

Note: The direction of magnetic field inside the solenoid is parallel to its axis.

Question 7.

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer:

Given:

Current in wire A, I_A = 8.0 A

Current in wire B, I_B = 5.0 A

Distance between the conductors A and B, d = 4 cm

Length of conductor on which we have to calculate force, L = 10cm

Intuitively we can break the problem into two parts,

1) the magnetic field due to wire A, at a distance d.

2) Then we introduce a current carrying conductor B, at distance d and find the force on it due to field created by wire A.

1) Field at distance d due to current I_A in conductor A is given by,

 …(1)

Where,

B = Magnetic field strength

I = current through the coil

�₀ is the permeability of free space.

�₀ = 4 × π × 10^-7 TmA^-1

d = distance of point P

By plugging the values in the equation (1), we get

⇒ |B| = 3.18 × 10^-6T

2) Force on a current carrying conductor due to a magnetic field is given by

F = B × I_B × L …(2)

Now by putting the values in equation (2) we get,

F = 3.18 × 10^-6T × 5A × 0.1 m

⇒ F = 2 × 10^-5N

So, the force on the 10 cm section on wire A is 2 × 10^-5N. Since the current is flowing in the same direction the force will be attractive in nature.

Note: The force will be same on both the wires, we can use Newton’s third law of motion to such conclusion.

Question 8.

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Answer:

Given:

Length of solenoid, L = 80cm

Number of turns = number of layers × number of turns per layer

Number of turns, n = 5 × 400 = 2000

Radius of solenoid, r = Diameter/2 = 0.9 cm

Current through the solenoid = 8.0A

We know that, magnetic field strength near the centre of solenoid is given by,

 …(1)

Where,

B = Magnetic field strength

n = total number of turns

I = current through the coil

�₀ is the permeability of free space.

�₀ = 4 × π × 10^-7 TmA^-1

r = radius of coil

Now, by plugging the values In equation (1), we get,

⇒ |B| = 2.512 × 10^-2T

Hence the magnetic field strength at the centre of the solenoid is 2.512 × 10^-2T.

Question 9.

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer:

Given:

Length of side of square, L = 10 cm

Number of turns, n = 20

Current through the square coil, I = 12 A

Angle between the normal to the coil and uniform magnetic field, θ = 30°

Magnitude of magnetic field, B = 0.80 T

The torque experienced by the coil in a magnetic field is given by,

T = n × B × I × A × sin(θ) …(1)

Where,

n = number of turns

B = Strength of magnetic field

I = Current through the coil

A = Area of cross-section of coil

A = L² = 0.1 × 0.1 = 0.01m² …(2)

θ = Angle between normal to cross-section of coil and magnetic field

Now by plugging the values in equation (1), we get

T = 20 × 0.80T × 12A × 0.01m² × sin30°

⇒ T = 0.96 Nm

∴ the magnitude torque experienced by the coil is 0.96 N-m.

Hence the torque experienced by the square coil is 0.96 Nm.

Question 10.

Two moving coil meters, M₁ and M₂ have the following particulars:

R₁ = 10Ω, N₁ = 30,

A₁ = 3.6 × 10^–3 m², B₁ = 0.25 T

R₂ = 14Ω, N₂ = 42,

A₂ = 1.8 × 10^–3 m², B₂ = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.

Answer:

Given:

For moving coil meter M₁

Resistance of wire, R₁ = 10Ω

Number of turns, N₁ = 30

Area of cross-section, A₁ = 3.6 × 10^-3 m²

Magnetic field strength, B₁ = 0.25 T

For moving coil meter M₂

Resistance of wire, R₂ = 14Ω

Number of turns, N₂ = 42

Area of cross-section, A₂ = 1.8 × 10^-3 m²

Magnetic field strength, B₂ = 0.50 T

Spring constant, K₁ = K₂ = K

a) Current sensitivity is given by,

For M₁,

 …(1)

By putting the values in equation 1, we have

⇒ 

For M₂

 …(2)

By putting the values in equation 2, we have

⇒ 

Now finding the ration of I₁ and I₂

⇒ 

Hence, the ratio of current sensitivities is 1.4.

b) Voltage sensitivity is given by,

For M₁,

⇒  …(3)

For M₂

⇒  …(4)

By dividing equation 3 by 4, we get

⇒ 

Hence, the ratio of voltage sensitivity of M₁ and M₂ is 1.

Question 11.

In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10^–19 C, m_e = 9.1 × 10^–31 kg)

Answer:

Given:

Magnetic field strength, B = 6.5 G = 6.5 × 10^-4T

Initial velocity of electron = 4.8 × 10⁶ ms^-1

Angle between the initial velocity of electron and magnetic field, θ = 90⁰

Force on the electron,

F_e = e × V × B × sinθ …(1)

Where,

e = charge on electron

V = velocity of electron

B = Magnetic field strength

θ = angle between direction of velocity and magnetic field

By putting the values in equation (1), we get

⇒ F_e = 1.6 × 10^-19 C × 4.8 × 10⁶ ms^-1 × 6.5 × 10^-4T × sin 90

⇒ F_e = 4.99 × 10^-16N

This force serves as the centripetal force, which explains the circular trajectory of the electron.

Centripetal force F_c = mv²/r …(2)

By equating equation (1) and equation (2) we get,

⇒ 

⇒ 

r = 4.2 cm.

Hence the radius of the path of electron shot into the magnetic field is 4.2 cm.

Question 12.

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer:

Given:

Magnetic field strength, B = 6.5 G = 6.5 × 10^-4 T

Initial velocity of electron = 4.8 × 10⁶ ms^-1

Angle between the initial velocity of electron and magnetic field, θ = 90⁰

We can relate the velocity of the electron to its angular frequency by the relation,

V = rω …(1)

Where,

V = velocity of electron

r = radius of path

ω = angular frequency

We understand that, magnetic force on the electron is equal to the centripetal force on it, hence we can write,

F_e = F_c

e × V × B = mV²/r …(2)

From equation (2) we can write,

⇒ eB = mV/r …(3)

Now, by putting value of V from equation (1) in equation (3)

⇒ e × B = m × ω …(4)

We know that, ω = 2πν

Putting in equation (4), we have

 …(5)

Now, by putting the values in equation (5) we get,

⇒ 

⇒ v = 18.2 × 10⁶Hz

Hence, the frequency of rotation is 18.2 × 10⁶Hz.

Note: The frequency of rotation is independent of the initial velocity of the electron.

Question 13.

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Answer:

Given:

Number of turns in the coil, n = 30

Radius of coil, r = 8 cm

Current through the coil, I = 6.0 A

Strength of magnetic field = 1.0 T

Angle between the direction of field and normal to coil, θ = 60°

We can understand that the counter torque required to prevent the coil from rotating is equal to the torque being applied by the magnetic field.

Torque on the coil due to magnetic field is given by,

T = n × B × I × A × sinθ …(1)

Where,

n = number of turns

B = Strength of magnetic field

I = Current through the coil

A = Area of cross-section of coil

A = πr² = 3.14 × (0.08 × 0.08) = 0.0201m² …(2)

θ = Angle between normal to cross-section of coil and magnetic field

Now, by putting the values in equation (1) we get,

⇒ T = 30 × 6.0T × 1A × 0.0201m² × sin60°

T = 3.133 Nm

Hence, the counter torque required to prevent the coil from rotating is 3.133 Nm.

Question 14.

Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer:

From equation (1) we can understand that, torques depends on the total area of cross-section and has no relation with the geometry of cross-section. Hence, the answer will remain unaltered if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area.

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Additional Exercises

Question 1.

Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Answer:

Here we have to find total magnetic field produced by the system so we will first find magnetic field due to each coil with direction and then add them in accordance with vector addition. Using the Right-hand thumb rule we can predict the direction of induced magnetic field in both the coils.

The orientation of both the coils is shown below in the figure.

The direction of induced magnetic field in both the coil can be found with the help of Right Hand Thumb Rule.

As Shown in figure.

So magnetic field in Coil X would be towards East, and in Coil Y would be towards West,

Explanation: if we curl our fingers in the Anticlockwise Sense in as in case of coil x our thumb will show the direction of the magnetic field which is towards East and if we curl our fingers in the clockwise Sense in as in case of coil Y our thumb will show the direction of the magnetic field which is towards West

The magnitude of magnetic field in a coil is given by the formula 

Where, B is the magnetic field

I is the current in the coil

R is the radius of coil in metres

N is the number of turns

 is the permittivity of free space = 4π × 10^-7 TmA^-1

For coil X

Current in the coil, I_x = 16A

Radius of the coil, R_x = 16cm = 0.16m

No. of turns of Coil, N �_x = 20

So magnetic field due to coil X, B_x = 

= 4π × 10^-4 T

= 1.2510^-3 T (Towards East)

For coil Y

Current in the coil, I_y = 18A

Radius of the coil, R_y = 10cm = 0.10m

No. of turns of Coil, N �_y = 25

So magnetic field due to coil Y, 

= 9π × 10^-4 T

= 2.82 × 10^-3 T (Towards West)

So, the Resultant magnetic field of the system = B_y- B_x

(since fields are in opposite directions)

= 2.82 × 10^-3 T- 1.2510^-3 T

= 1.57 × 10^-3 T (Towards West)

So net magnetic field of the system is B = 1.57 × 10^-3 T, Towards West direction

Question 2.

A magnetic field of 100 G (1 G = 10^–4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^–3 m². The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^–1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

Answer:

Here, we have a particular value of No. of turns per unit Length and Current in the coil in order to obtain the given magnetic field.

The Required Magnetic field B = 100 G = 100 × 10^–4= 10^–2 T

Maximum Number of turns per unit length, n = 1000/m

Maximum Current flowing in the coil, I = 15 A

Permeability of free space, μ₀ = 4π × 10^–4TmA^-1

We know magnetic field for a solenoid is given by

B = 𝜇 �₀nI

Where,

B is the magnetic field

n is the Number of turns of the coil per unit length

I is the current in the coil

𝜇 _�0 is the permittivity of free space

Explanation: here we can change the number of turns per unit length n (maximum 1000 turns per m) and current I (maximum 15A) since 𝜇_ois a constant and we have to obtain a fixed value of magnetic field B = 10^-2T, so we will alter values of n and I accordingly

Using the formula B = 𝜇 �₀nI

We get, B/𝜇 �₀ = nI

So, 

= 7957.74 A/m ≈ 8000 A/m

If we take number of turns of coil N = 400,

If we take length of the coil L = 50 cm = 0.5m,

No. of turns per unit length, n = N/L= 400/0.5 = 800/m

(less than 1000/m)

If we take current I = 10 A

(less than 15 A),

We get nI = 8000 A/m

Area of Coil A = 10^-3 m² = πr², where r is radius of coil

We get 

⇒ 

⇒ r = 0.017m

The values of N, L and I are within the proposed limits, Though there can be other values as well.

Question 3.

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,



(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

, approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Answer:

We know for a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

𝜇 _�0 is the permittivity of free space

As can be seen in the figure.

(a) To find the magnitude of magnetic field B at the centre of a circular coil.

Distance from centre, x = 0 m (as we are finding magnetic field at the centre of coil)

No. of turns of coil N = 1

Putting these values in equation ,

we get,

Or 

Or 

This is same as the known result for magnetic field of a current carrying loop at the centre.

(b) The situation has been represented in the figure, suppose two circular loops of equal Radius R carry equal current I in same direction, and are also separated by a distance R,

Now suppose a point at a distance d from midpoint of separation of the coils,so distance from one coil, 

And distance from second coil, x₁

Magnetic field due to 1st coil is given by 

Magnetic field due to 2nd coil is given by 

Net magnetic field due to both the coils is given by B = B₁ B₂

Explanation: Using Right hand thumb rule if curl our fingers in the direction of current the thumb shows the direction of the magnetic field,we find that magnetic field due to both coil is in same direction so their magnitude can be simply added to give net magnetic field

So we get,

Putting the values of x₁ and x₂ we get

i.e.

i.e. 

Expanding equations of second order we get

i.e.

Since d is very small compared to R so neglecting  and taking terms of R² common from both the terms we get

Explanation: we have used the binomial expansions

and

Where x = 4d/5R and we have neglected second terms onwards because d<<R

So d²/R² ≈ 0

i.e. 

So 

Solving and putting value of  = 0.72

We get

Hence proved the required result

Question 4.

A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound.

If the current in the wire is 11 A, what is the magnetic field

(a) outside the toroid,

(b) inside the core of the toroid, and

(c) in the empty space surrounded by the toroid.

Answer:

The Toroid is same as a solenoid but is bound in a circular region as compared to straight cylindrical region in a solenoid the length of a toroid is given by L = 2πR, (Perimeter of a Circle)

where R is mean radius of cross section of toroid

 = 25.5 cm = 0.255 m

here Total number of turns of toroid N = 3500,

no. of turns per unit length n = N/L =  = 2475.74/m

the current flowing in Toroid I = 11 A

The toroid has been shown in the following figure

(a) Since Toroid is same as a solenoid so magnetic field outside the toroid = 0 T same as in case of solenoid.

Explanation: if we apply Ampere Circuital Lawto find magnetic field B at small element dl due to a current carrying conductor with current I flowing through it andis permittivity of free space, outside the solenoid or toroid, where there is no conductor/wire/current carrying loop, the current I = 0 so we getSo Integrating over length l we get Bl = 0 so magnetic field B = 0 i.e. at all places where there is no current carrying loop magnetic field B = 0

(b) The magnetic field in the inner core of toroid is given by

B = 𝜇onI

Where n is no. of turns per unit length,

I is the current flowing in Toroid,

𝜇o is the permittivity of free space =  TmA^-1

So putting all values, B = = T

i.e. Magnetic field inside core is 0.03 T.

(c) Again it’s the same case as of exterior of a solenoid as explained above so magnetic field is Zero, here also in the empty space between the toroid.

Question 5.

Answer the following questions:

A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

Answer:

When a charged particle enters a magnetic field it experiences a force known as Magnetic Lorentz Force which is given by:

Where  is the force on the Charged particle having charge q,moving with a velocity  in a magnetic Field 

Note:,andare vector quantities, and charge q is a scalar quantity, andis the cross product or vector product ofand,so resultant is perpendicular to plane containingand

if we evaluate the formula we get,

Where,  and are magnitude of velocity and magnetic field respectively, θ is the angle between Velocity of particle and magnetic field and ŵ is a unit vector perpendicular to plane containing  and 

When angle between Velocity of particle and magnetic field  is either  or  i.e. particle is either moving parallel or antiparallel to the field we can see force on the particle,

F = 0 N, since sinθ = 0 when θ is either 0⁰ or 180⁰,

The particle came out of the magnetic field un-deflected, and with constant velocity i.e. it did not experience any force so we can say that particle was moving in same direction as that of magnetic field i.e. from East Towards West or in opposite direction from West towards East and magnitude of velocity can be any value.

The following figure depicts both the situations

Question 6.

Answer the following questions:

A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

Answer:

When a charged particle enters a magnetic field it experiences a force known as Magnetic Lorentz Force which is given by

→ 

Now, due to magnetic lorentz force, the particle accelerates so the velocity of particle changes, Magnetic field changes is both magnitude and direction so the Magnitude of Force on particle also keeps on changing, hence the velocity of particle also changes.

Note: Acceleration is Rate of Change of Velocity, so if a particle experiences force, it accelerates and hence its velocity changes. Velocity is a vector quantity so it depends on direction, so velocity of a particle changes if its magnitude or speed changes, direction remain the same, or velocity changes when direction of particle changes and magnitude of velocity or speed remains same, or velocity changes when both speed and direction changes.

In this Case force on the particle is always perpendicular to its velocity, which only changes the direction of the particle and its speed remains same, i.e. velocity of particle changes due to change in direction of particle and magnitude of velocity or speed same as it entered.

Following Figure depict the direction of force on particle at any instant

Question 7.

Answer the following questions:

An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Answer:

If the Charged particle have to go undeflected then there should be no force experienced by it, or resultant of all the forces on the particle is zero. Here, the charged particle is experiencing Magnetic Lorentz force and electrostatic force due to electric field

We know Magnetic Lorentz Force is given by

Direction of the Magnetic force is perpendicular to plane containing V and B. and for a positively charged particle can be found out with help of Fleming’s left hand thumb rule.

Electrostatic force is given by

F = qE

Where F is the force experienced by particle having charge q and moving in an electric field E, direction of force is same as that of Electric field for a positively charged particle and is exactly reversed in case of negatively charged particle.

So both the forces on particle should be equal in magnitude and opposite in direction.

Since Electric Field is from North to South, and particle is an electron i.e. negatively charged so force on it should be towards North so force due to magnetic field should be towards south .As shown in figure

Using the Fleming left hand rule we will decide the direction of magnetic field

Flemings rule : If the stretch our forefinger middle finger and thumb mutually perpendicular to each other then if forefinger depicts the direction of magnetic field, middle finger depicts the direction of the current(Direction of Velocity of Positive Charge) then, Thumb depicts the direction of Force.

Applying the rule

Since Electron is negatively charged and moving from West to east so direction of current is from east to west, depicted by middle finger, force should be towards both depicted by our thumb, so we find our forefinger depicting magnetic field is in a direction vertically downwards in the plane

Now equating magnitude of both the forces

Here θ = 90° because Velocity of particle is perpendicular to Magnetic field,

So,sinθ = 1

i.e. 

or magnitude of magnetic field is given by

, in a direction vertically downwards.

Question 8.

An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field

(a) is transverse to its initial velocity,

(b) makes an angle of 30° with the initial velocity.

Answer:

When an charged particle having charge q is accelerated by a potential difference V volts it gain kinetic energy given by

K = Vq, which imparts it with velocity

We know kinetic energy of a Particle with mass m kg and velocity v is given by

Equating both equations

We get velocity of the particle as

Here Potential difference of the particle, V = 2.0 kV

= 2000 V

Particle is electron, charge of electron

q = 1.6 × 10 ^-19 C

mass of electron

m = 9.1 × 10^-31 Kg

so velocity of the particle,

m/s

Solving we get

Velocity of particle,

v = m/s

(a) Now here the particle enters the magnetic field orthogonally so it will experience Lorentz magnetic force is given by :

Where  is the force on the Charged particle having charge q, moving with a velocity  in a magnetic Field 

Where,  and are magnitude of velocity and magnetic field respectively, θ is the angle between Velocity of particle and magnetic field and ŵ is a unit vector perpendicular to plane containing  and 

Here, since particle entered orthogonally now force on particle will be perpendicular to both magnetic field and its velocity at every instant, since magnetic field and velocity are in one plane and force is in a plane perpendicular to them, this force does not change the magnitude of velocity or speed it only changed it the direction of particle and force always act tangentially to the velocity of the particle causing it to move in a circular trajectory

As shown in the figure

Now the required centripetal force for circular motion is provided by the magnetic force we know centripetal force is given by

,

Where F is the centripetal force,

m is the mass of the particle,

moving in a circular region of radius r,

with velocity v

Magnitude of magnetic force on particle moving with velocity v in a magnetic field B orthogonally is

F = qvB(sinθ)

here  at every instant so

F = qvB

Equating both equations

We get radius trajectory of the particle,

The particle is electron, thus, charge of electron is:

q = 1.6 × 10 ^-19 C

mass of electron

 Kg

Velocity of particle,

v = m/s

Magnitude of magnetic field

B = 0.15 T

so 

= 100.47 m

= 1.00 mm

So trajectory is a circle of radius 1.00 mm normal to plane of magnetic field

(b) when the particle enters the magnetic field, with its velocity at some angle with it, and not exactly orthogonal, particle follows a helical path, since there is force on the component of velocity which is perpendicular to the magnetic field, and No force is experience due to the component of velocity parallel to magnetic field, which keeps moving the particle forward so path is helical instead of circular. Velocity makes angle of 30 degree with the magnetic field, consider magnetic field along x axis

so particle moving in x-y plane initially making an angle 30^o with x axis, so using the Fleming left hand rule we find out the force is Toward positive Z axis,

as shown in figure

Explanation: It can be visualised as particle moving in circular path, at the same time moving forward, like an spiral, if there would have been no horizontal component of velocity it would have moved in a circular path.

Now particle tends to move in circular path in y-z plane due to Y component of velocity perpendicular to the field,and move forward along x axis due to x component of velocity,

Now angle made by velocity with x-axis,

θ = 30°

So x component of velocity,

 m/s

So y component of velocity,

 m/s

For the radius of the path we will again use the equation

where, q is charge of Particle which is electron so

C

m is mass of particle, i.e. mass of electron

Kg

Component of Velocity of particle undergoing circular motion,

v_y = m/s

Magnitude of magnetic field

B = 0.15 T

Putting values in equation 

m

= 0.5mm

So radius of the helical path followed by electron is 0.5 mm and is moving along magnetic field with speed m/s.

Question 9.

A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10⁵ V m^–1, make a simple guess as to what the beam contains. Why is the answer not unique?

Answer:

Now here the charged particle is two forces, magnetic Lorentz force due to magnetic field and Electrostatic force and electrostatic force due to electric field, now for particle to go undeflected, the direction of both the forces should be exactly opposite to each other, and magnitude should be equal. direction of force is same as that of Electric field for a positively charged particle and is exactly reversed in case of negatively charged particle and direction of magnetic force is given by Flemings left hand rule.

The situation is depicted in the figue.

Electrostatic force is given by

Where F is the force experienced by particle having charge q and moving in an electric field E Magnitude of magnetic force on particle moving with velocity v in a magnetic field B, velocity makes angle  with the Field

here ,since velocity is perpendicular to magnetic field

so 

So Equating both we get

so we get

velocity of the particle, 

now the particle is accelerated through a potential difference When an charged particle having charge q is accelerated by a potential difference V volts it gain kinetic energy given by

K = Vq, which imparts it with velocity

We know kinetic energy of a Particle with mass m and velocity v is given by

Equating both equations

We get velocity of the particle as

So Equating both the values of velocity v we get

Solving we get

Here we are given,

Electric field E = 9.0 × 10⁵ V m^–1

Magnetic Field B = 0.75 T

The Potential difference V = 15 kV = 15000 V

Putting all the values we get

⇒ q/m = 4.8 10⁷ C/Kg

This is the charge to mass ratio of Deuterium ions Dor deuterons, which is a stable isotope of Hydrogen having one proton and one neutron in its nucleus.

So charge on Deuterium ion = e (since 1 charge)

= C

Mass of Deuterium ion,

m = Kg

so charge to mass ratio,

 ≈ 4.8 × 10⁷ C/Kg

the answer is not unique because only the ratio of charge to mass is determined. There can Other possible species whose charge to mass ratio is same as this value examples are

He^, Li^, etc.

Question 10.

A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s^–2.

Answer:

Given:

Length of rod = 0.45 m

Mass of rod = 60 gm

Current = 5.0 A

Acceleration due to gravity, g = 9.8 m/s²

(a) Let magnetic field in conductor so that tension is zero = B

Tension will be zero if the force due to magnetic field and the weight is balanced.

We know that,

B × I × l = m × g …(1)

Where,

B = magnetic field

I = current through the conductor

l = length of conductor

m = mass of conductor

g = acceleration due to gravity

From equation (1) we have,

Now plugging the values of m, g, I and l in equation (1)

⇒ 

⇒ B = 0.26 T

So, a magnetic field of 0.26 Tesla has to be set up normal to the conductor in order to get zero tension in the supporting wire. Using Fleming’s left hand rule we find that the magnetic field applies a force in upward direction on the conductor which balances its weight.

(b) Let total tension in wire if the direction of current is reversed = P

According to question, the magnetic field is kept the same. Using Fleming’s left hand rule we find that Force due to magnetic field on the conductor acts downwards.

Writing the balancing equations:

P = B × I × l m × g …(2)

Where,

P = tension

B = magnetic field

I = current through the conductor

l = length of conductor

m = mass of conductor

g = acceleration due to gravity

Now, putting the values in equation (2) we get,

⇒ P = 0.26T × 5A × 0.45A 0.06Kg × 9.8ms^-2

⇒ P = 1.17 N

The tension in wire is 1.17 N.

Question 11.

The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Answer:

Given:

Current through the wire, I = 300 A

Length of wire, L₁ = 70 cm

Length of wire, L₂ = 70 cm

Distance between the wires, d = 1.5 cm

We understand that current carrying conductors create magnetic fields which in turn applies force on nearby current carrying conductor(s).

We know that,

 …(1)

Where,

F = force applied

�₀ = permeability of free space

I = current through the conductors

d = separation between the conductors

Now, putting the values of F, �₀, I and d in equation (1)

⇒ 

⇒ F = 1.2 Nm^-1

Hence, the force applied on the wires is 1.2 N per metre. The force is repulsive in nature because the current is carried in the opposite direction.

Question 12.

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

(a) the wire intersects the axis,

(b) the wire is turned from N-S to northeast-northwest direction,

(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Answer:

Given:

Magnetic field strength, B = 1.5 T

Radius of cylinder, r = 10 cm

Current in wire travelling north to south, I = 7A

(a) Let the force on wire = F₁

If the wire intersects the axis of the magnetic field, then we observe that length of wire will be equal to the diameter of the cylinder.

L₁ = 2 × r = 20 cm

We know that,

F = B × I × L₁ × sinθ …(1)

Where,

F = Force on wire

B = magnetic field strength

I = current through the wire

L₁ = length of wire

θ = angle between direction of field and direction of current

Since, the field is along east to west and current flows along north south, direction so, θ = 90°.

By putting the values of B, I and L₁ in equation 1.

⇒ F = 1.5T × 7A × 0.2m × sin90°

⇒ F = 2.1 N

So, force on wire is 2.1 N and it acts in vertically downward direction.

(b) Let the force on wire = F₂

If the wire is turned towards Northeast-Northwest direction.

We observe that wire has been turned 45°, i.e. θ = 45°

L₂ = L₁/sinθ.

L₂ = 2 × L₁

F₂ = B × I × L₂ × sinθ

F₂ = B × I × L₂

F₂ = 2.1 N

Hence the force is 2.1 N and it acts vertically downwards.

(c) Let, the force on wire = F₃

The wire has been lowered 6 cm,

⇒ x = ( r²-p²)^0.5

⇒ x = ( 100-36)^0.5

⇒ x = 8

So length of wire in the magnetic field region = 16 cm

(By using the geometry)

We know that,

F = B × I × L₁ × sinθ …(1)

Now, putting the values in B, T and I in equation 1.

⇒ F = 7T × 0.16A × 1.5T

⇒ F = 1.68 N

The direction of the force will be vertically downward, the direction of force is given by screw rule.

Question 13.

A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Answer:

Given:

Magnetic field strength, B = 3000G

B = 3000 × 10^-4T

B = 0.3 T

Length of rectangular loop = 10 cm

Width of loop = 5 cm

Current in loop = 12 A

We know that,

Torque, T = IA × B …(1)

Where, I = current through loop

A = area of cross-section

B = magnetic field strength

Note: We take vector normal to the cross-section for A.

For each of the case we plug in the variables in equation (1).

(a) T = 12A × (50 × 10^-4) î m² × 0.3 T

 = -1.8 × 10^-2 ĵ Nm (∵ î ×  = -ĵ )

Force on the loop is Zero as the angle between area of cross section and magnetic field is Zero.

(b) This case is similar to A, hence the results will also be same. There is no change in any vectors.

(c) T = IA × B

We observe that A is perpendicular to x-z plane and B is along z axis.

⇒ T = -12A × (50 × 10^-4)m² ĵ × 0.3T 

⇒ T = -1.8 × 10^-2 î Nm (∵ ĵ ×  = -î )

(d) Magnitude of torque,

|T| = IA × B

⇒ |T| = 12 × (50 × 10^-4)m² × 0.3T

⇒ |T| = 1.8 × 10^-2 Nm

This torque is makes 240° with the positive x direction. The force again is Zero.

(e) T = IA × B

 = 12 × (50 × 10^-4) m² × 0.3 T

 = 0

Since cross section and magnetic field are in same direction net torque is zero. Net force is also Zero.

(f) 

 = 12 × (50 × 10^-4)  × 0.3 

 = 0

Hence the torque and forces are zero.

Stable equilibrium:

In case E, the angle between A and B is zero. If we displace the wire, it will come back in this position, hence it is the stable equilibrium condition.

Unstable equilibrium:

In case F, the angle between A and B is 180°. If the wire is displaced, it will not come back in this position so we can conclude that it is the case for unstable equilibrium.

Question 14.

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the

(a) total torque on the coil,

(b) total force on the coil,

(c) average force on each electron in the coil due to the magnetic field?

(The coil is made of copper wire of cross-sectional area 10^–5 m², and the free electron density in copper is given to be about 10²⁹ m^–3.)

Answer:

Given:

Number of turns, N = 20

Radius of coil, r = 10cm

Magnetic field strength = 0.10 T

Current in the coil = 5 A

a) The torque on the coil is zero because the magnetic field is uniform along the coil.

b) The total force on the coil is Zero due to uniform magnetic field.

c) Let average force on each electron in the coil due to magnetic field = F

Cross-sectional area of the wire, A = 10^-5 m²

Free electron density or number of free electron per metre cube of copper, D = 10²⁹

Charge in each electron = 1.6 × 10¹⁹ C

We know that,

 …(1)

Where,

F = force due to magnetic field

B = Magnetic field strength

e = charge on electron

V_d = drift velocity of electron

V_d = I/ (NeA) …(2)

Where, I = current through the coil

N = Density of free electron

A = cross sectional area.

Putting values in equation (2)

⇒ 

⇒ V_d = 3.125 × 10⁴³ ms^-1

F = B × e × V_d …(3)

By putting the values of B, e and V_d in equation 3, we get,

⇒ 

⇒ F = 5 × 10^-25 N

Hence, average force on each electron is 5 × 10^-25 N.

Question 15.

A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^–2.

Answer:

Given:

Length of solenoid, L₁ = 60 cm

Radius of solenoid, r = 4 cm

Number of layers, n₁ = 3

Number of turns, N = 300

Total number of turns,n = N × n₁ = 900

Length of wire, L₂ = 2 cm

Mass of wire, m = 2.5 gm

Current flowing through the wire, I₂ = 6 A

Acceleration due to gravity, g = 9.8 ms^-2

Intuitively, this problem can be broken down into three parts. In first part we will establish the magnetic field inside the solenoid and in second part we introduce a current carrying conductor in the magnetic field. Then we can evaluate the force on the wire. In the final part we try to find the balancing force and finally the current.

Part (1)

We know that magnetic field inside a solenoid is given by,

 …(1)

Where,

B = Magnetic field strength

n = total number of turns

I₁ = current through the coil

L₁ = length of the coil

�₀ is the permeability of free space.

�₀ = 4 × π × 10^-7 TmA^-1

Part (2)

We know that, when a current carrying conductor is placed in a magnetic field, it experiences a force given by,

F = B × I₂ × L₂ …(2)

Now putting the value of B from equation 1 into equation 2.

 …(3)

Where, I₂ = current through the conductor

L₂ = length of wire

Part (3)

Since, the wire is suspended inside the solenoid, the upward force on it must be equal to its weight.

By equating the weight (m × g) of the body with force in equation (3), we get

m × g =  …(4)

⇒ I₁ = 

⇒ I₁ = 

Note: It can be noted that in SI base units Tesla(T) = Kgs^-2A^-1

⇒ I₁ = 108A

Hence the current flowing through the solenoid is 108A.

for the suspension of the wire.

Question 16.

A galvanometer coil has a resistance of 12 and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Answer:

Given:

Resistance of the galvanometer coil, G = 12Ω

Current for full scale deflection = 3mA

For converting a galvanometer into a Voltmeter we can connect a resistance in series with the galvanometer. A minimum value of current should be allowed to flow through the voltmeter.

Required range of the voltmeter = 0 to 18 V

V = 18

Let the resistance needed = R_s

 …(1)

By plugging the values of V, G and I_s in equation 1, we get,

⇒

⇒ R_s = 5988 Ω

For converting the galvanometer into a voltmeter of range 0-18 we can add a resistance of 5988Ω, is series with the galvanometer.

Question 17.

A galvanometer coil has a resistance of 15 and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Answer:

Given:

Resistance of galvanometer coil, G = 15 Ω

Current for full scale deflection, I_g = 4mA

Upper limit of ammeter I = 6A

A galvanometer can detect only small currents and their direction. Thus, to measure large currents it is converted into an ammeter. A galvanometer can be converted into an ammeter by connecting a low resistance called shunt resistance in parallel to the galvanometer

Let shunt resistance = S

We know that value of shunt is given by:

 …(1)

By plugging the values of I_g, I and G into Equation (1), we get,

⇒ 

⇒ S = 10mΩ

For converting the galvanometer into an ammeter of range 0-6 we can add a shunt of resistance 10mΩ.