Moving Charges And MagnetismClass 12 Physics NCERT Solutions
13 Solutions
Generated by KedovoAI
Solution 1 of 13
Q1EXERCISES
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?
Solution
The magnitude of the magnetic field at the centre of a circular coil with N turns is given by the formula:
B = (μ₀ * N * I) / (2 * R)
Where:
- N = number of turns = 100
- I = current = 0.40 A
- R = radius of the coil = 8.0 cm = 0.08 m
- μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
Substituting the given values into the formula:
B = (4π × 10⁻⁷ T m A⁻¹ × 100 × 0.40 A) / (2 × 0.08 m)
B = (4π × 10⁻⁷ × 40) / 0.16 T
B = (160π × 10⁻⁷) / 0.16 T
B = 1000π × 10⁻⁷ T
B = π × 10⁻⁴ T
B ≈ 3.14 × 10⁻⁴ T
Therefore, the magnitude of the magnetic field at the centre of the coil is 3.14 × 10⁻⁴ T.
Q2EXERCISES
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?
Solution
The magnitude of the magnetic field (B) at a distance (r) from a long straight wire carrying a current (I) is given by Ampere's circuital law:
B = (μ₀ * I) / (2 * π * r)
Where:
- I = current = 35 A
- r = distance from the wire = 20 cm = 0.20 m
- μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
Substituting the values:
B = (4π × 10⁻⁷ T m A⁻¹ × 35 A) / (2 * π * 0.20 m)
B = (2 × 10⁻⁷ × 35) / 0.20 T
B = (70 × 10⁻⁷) / 0.20 T
B = 350 × 10⁻⁷ T
B = 3.5 × 10⁻⁵ T
Thus, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10⁻⁵ T.
Q3EXERCISES
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.
Solution
Magnitude of the Magnetic Field
The magnitude of the magnetic field (B) at a distance (r) from a long straight wire is given by:
B = (μ₀ * I) / (2 * π * r)
Where:
- I = current = 50 A
- r = distance from the wire = 2.5 m
- μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
Substituting the values:
B = (4π × 10⁻⁷ T m A⁻¹ × 50 A) / (2 * π * 2.5 m)
B = (2 × 10⁻⁷ × 50) / 2.5 T
B = (100 × 10⁻⁷) / 2.5 T
B = 40 × 10⁻⁷ T
B = 4 × 10⁻⁶ T
Direction of the Magnetic Field
The direction of the magnetic field can be determined using the Right-Hand Thumb Rule.
- Point the thumb of your right hand in the direction of the current, which is from north to south.
- Curl your fingers around the wire.
- At a point 2.5 m east of the wire, your fingers will point vertically upwards.
Therefore, the magnitude of the magnetic field is 4 × 10⁻⁶ T, and its direction is vertically upwards.
Q4EXERCISES
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?
Solution
Magnitude of the Magnetic Field
The magnitude of the magnetic field (B) at a distance (r) from a long straight wire is given by:
B = (μ₀ * I) / (2 * π * r)
Where:
- I = current = 90 A
- r = distance from the wire = 1.5 m
- μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
Substituting the values:
B = (4π × 10⁻⁷ T m A⁻¹ × 90 A) / (2 * π * 1.5 m)
B = (2 × 10⁻⁷ × 90) / 1.5 T
B = (180 × 10⁻⁷) / 1.5 T
B = 120 × 10⁻⁷ T
B = 1.2 × 10⁻⁵ T
Direction of the Magnetic Field
The direction of the magnetic field is determined using the Right-Hand Thumb Rule.
- Point the thumb of your right hand in the direction of the current, which is from east to west.
- Curl your fingers around the wire.
- At a point 1.5 m below the wire, your fingers will point towards the south.
Therefore, the magnitude of the magnetic field is 1.2 × 10⁻⁵ T, and its direction is towards the south.
Q5EXERCISES
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?
Solution
The magnetic force (F) on a straight wire of length (l) carrying a current (I) in a uniform magnetic field (B) is given by:
F = I * l * B * sin(θ)
Where θ is the angle between the direction of the current and the magnetic field.
To find the force per unit length (f), we divide the force F by the length l:
f = F / l = I * B * sin(θ)
Given:
- I = current = 8 A
- B = magnetic field = 0.15 T
- θ = angle = 30°
Substituting the values into the formula for force per unit length:
f = 8 A × 0.15 T × sin(30°)
Since sin(30°) = 0.5:
f = 8 × 0.15 × 0.5 N/m
f = 8 × 0.075 N/m
f = 0.6 N/m
Therefore, the magnitude of the magnetic force per unit length on the wire is 0.6 N/m.
Q6EXERCISES
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?
Solution
The magnetic force (F) on a straight wire of length (l) carrying a current (I) in a uniform magnetic field (B) is given by the formula:
F = I * l * B * sin(θ)
Where θ is the angle between the wire (direction of current) and the magnetic field.
Given:
- l = length of the wire = 3.0 cm = 0.03 m
- I = current = 10 A
- B = magnetic field inside the solenoid = 0.27 T
The magnetic field inside a solenoid is directed along its axis. The wire is placed perpendicular to the axis, so the angle θ between the current and the magnetic field is 90°.
- θ = 90°
- sin(90°) = 1
Substituting the values into the formula:
F = 10 A × 0.03 m × 0.27 T × sin(90°)
F = 10 × 0.03 × 0.27 × 1 N
F = 0.3 × 0.27 N
F = 0.081 N
Therefore, the magnetic force on the wire is 0.081 N.
Q7EXERCISES
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.
Solution
The force per unit length (f) between two long parallel wires is given by:
f = (μ₀ * Iₐ * Iₑ) / (2 * π * d)
Where:
- Iₐ = current in wire A = 8.0 A
- Iₑ = current in wire B = 5.0 A
- d = distance between the wires = 4.0 cm = 0.04 m
- μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
First, we calculate the force per unit length:
f = (4π × 10⁻⁷ T m A⁻¹ × 8.0 A × 5.0 A) / (2 * π * 0.04 m)
f = (2 × 10⁻⁷ × 8.0 × 5.0) / 0.04 N/m
f = (2 × 10⁻⁷ × 40) / 0.04 N/m
f = (80 × 10⁻⁷) / 0.04 N/m
f = 2000 × 10⁻⁷ N/m
f = 2.0 × 10⁻⁴ N/m
Now, we need to find the total force (F) on a 10 cm section of wire A.
The length of the section is L = 10 cm = 0.1 m.
F = f × L
F = (2.0 × 10⁻⁴ N/m) × 0.1 m
F = 2.0 × 10⁻⁵ N
Direction of the force: Since the currents in both wires are in the same direction, the force between them is attractive. Therefore, the force on wire A is directed towards wire B.
The estimated force on the 10 cm section of wire A is 2.0 × 10⁻⁵ N, and it is attractive (towards wire B).
Q8EXERCISES
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.
Solution
The magnitude of the magnetic field (B) inside a long solenoid near its centre is given by the formula:
B = μ₀ * n * I
Where:
- μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
- I = current = 8.0 A
- n = number of turns per unit length
First, we need to calculate 'n'.
- Length of the solenoid, L = 80 cm = 0.8 m
- Number of layers = 5
- Number of turns per layer = 400
- Total number of turns, N = 5 layers × 400 turns/layer = 2000 turns
Now, we can find the number of turns per unit length (n):
n = N / L
n = 2000 turns / 0.8 m
n = 2500 turns/m
Now, we can calculate the magnetic field B:
B = (4π × 10⁻⁷ T m A⁻¹) × (2500 m⁻¹) × (8.0 A)
B = 4π × 10⁻⁷ × 2500 × 8.0 T
B = 4π × 10⁻⁷ × 20000 T
B = 8π × 10⁻³ T
B ≈ 8 × 3.14 × 10⁻³ T
B ≈ 25.12 × 10⁻³ T
B ≈ 2.5 × 10⁻² T
(Note: The diameter of the solenoid is not needed for this calculation as long as the solenoid is considered long, i.e., its length is much greater than its diameter).
Therefore, the magnitude of the magnetic field inside the solenoid near its centre is approximately 2.5 × 10⁻² T.
Q9EXERCISES
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?
Solution
The magnitude of the torque (τ) experienced by a current-carrying coil in a uniform magnetic field is given by:
τ = N * I * A * B * sin(θ)
Where:
- N = number of turns = 20
- I = current = 12 A
- A = area of the coil
- B = magnitude of the magnetic field = 0.80 T
- θ = angle between the normal to the plane of the coil and the magnetic field = 30°
First, we calculate the area (A) of the square coil:
- Side of the square coil = 10 cm = 0.1 m
- Area, A = (side)² = (0.1 m)² = 0.01 m²
Now, we can substitute all the values into the torque formula:
τ = 20 × 12 A × 0.01 m² × 0.80 T × sin(30°)
Since sin(30°) = 0.5:
τ = 20 × 12 × 0.01 × 0.80 × 0.5 N m
τ = 240 × 0.01 × 0.80 × 0.5 N m
τ = 2.4 × 0.80 × 0.5 N m
τ = 1.92 × 0.5 N m
τ = 0.96 N m
Therefore, the magnitude of the torque experienced by the coil is 0.96 N m.
Q10EXERCISES
Two moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T R₂ = 14 Ω, 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.
Solution
Let k be the spring constant for both meters.
(a) Ratio of Current Sensitivity
The current sensitivity (CS) of a moving coil galvanometer is defined as the deflection per unit current (φ/I).
CS = (N * A * B) / k
For meter M₁:
CS₁ = (N₁ * A₁ * B₁) / k
For meter M₂:
CS₂ = (N₂ * A₂ * B₂) / k
The ratio of the current sensitivity of M₂ to M₁ is:
Ratio = CS₂ / CS₁ = [(N₂ * A₂ * B₂) / k] / [(N₁ * A₁ * B₁) / k]
Ratio = (N₂ * A₂ * B₂) / (N₁ * A₁ * B₁)
Substituting the given values:
Ratio = (42 × 1.8 × 10⁻³ × 0.50) / (30 × 3.6 × 10⁻³ × 0.25)
Ratio = (42 × 1.8 × 0.50) / (30 × 3.6 × 0.25)
Ratio = 37.8 / 27
Ratio = 1.4
So, the ratio of current sensitivity (M₂ to M₁) is 1.4.
(b) Ratio of Voltage Sensitivity
The voltage sensitivity (VS) is defined as the deflection per unit voltage (φ/V).
VS = (N * A * B) / (k * R)
For meter M₁:
VS₁ = (N₁ * A₁ * B₁) / (k * R₁)
For meter M₂:
VS₂ = (N₂ * A₂ * B₂) / (k * R₂)
The ratio of the voltage sensitivity of M₂ to M₁ is:
Ratio = VS₂ / VS₁ = [(N₂ * A₂ * B₂) / (k * R₂)] / [(N₁ * A₁ * B₁) / (k * R₁)]
Ratio = [(N₂ * A₂ * B₂) / (N₁ * A₁ * B₁)] × (R₁ / R₂)
We already calculated the first part of this expression in part (a), which is 1.4.
Ratio = 1.4 × (R₁ / R₂)
Ratio = 1.4 × (10 Ω / 14 Ω)
Ratio = 1.4 × (10 / 14)
Ratio = 1.4 × (5 / 7)
Ratio = (14/10) × (5/7)
Ratio = (2/2) × 1 = 1
So, the ratio of voltage sensitivity (M₂ to M₁) is 1.
Q11EXERCISES
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, mₑ = 9.1 × 10⁻³¹ kg)
Solution
Explanation of Circular Path
When an electron enters a magnetic field normal to its direction of motion, it experiences a magnetic force given by the Lorentz force equation: F = q(v × B).
- Direction of Force: According to the vector cross product, the direction of this force is always perpendicular to both the velocity vector (v) of the electron and the magnetic field vector (B).
- No Work Done: Since the force is always perpendicular to the direction of motion (velocity), no work is done on the electron by the magnetic field. Consequently, the kinetic energy and the speed of the electron remain constant.
- Centripetal Force: A constant force that is always perpendicular to the velocity of a particle acts as a centripetal force. This force continuously changes the direction of the electron's velocity without changing its magnitude, causing the electron to move in a circular path.
Determination of the Radius
The magnetic force provides the necessary centripetal force for the circular motion.
Magnetic Force = Centripetal Force
q * v * B * sin(θ) = (mₑ * v²) / r
Since the electron is shot normal to the field, θ = 90° and sin(90°) = 1.
So, e * v * B = (mₑ * v²) / r
Rearranging the formula to solve for the radius (r):
r = (mₑ * v) / (e * B)
Given:
- B = 6.5 G = 6.5 × 10⁻⁴ T
- v = 4.8 × 10⁶ m/s
- e = 1.5 × 10⁻¹⁹ C (as given in the question)
- mₑ = 9.1 × 10⁻³¹ kg
Substituting the values:
r = (9.1 × 10⁻³¹ kg × 4.8 × 10⁶ m/s) / (1.5 × 10⁻¹⁹ C × 6.5 × 10⁻⁴ T)
r = (43.68 × 10⁻²⁵) / (9.75 × 10⁻²³) m
r ≈ 4.48 × 10⁻² m
r ≈ 4.5 cm
Therefore, the radius of the circular orbit is approximately 4.5 cm.
Q12EXERCISES
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.
Solution
Calculation of Frequency
The frequency of revolution (ν) of a charged particle in a circular orbit within a magnetic field is also known as the cyclotron frequency. It is related to the angular frequency (ω) by ω = 2πν.
The magnetic force provides the centripetal force:
e * v * B = (mₑ * v²) / r
We know that for circular motion, v = ωr. Substituting this into the equation:
e * (ωr) * B = mₑ * (ωr)² / r
e * ω * r * B = mₑ * ω² * r
e * B = mₑ * ω
ω = (e * B) / mₑ
Since ω = 2πν, the frequency ν is:
ν = ω / (2π) = (e * B) / (2 * π * mₑ)
Using the values from Exercise 4.11:
- B = 6.5 × 10⁻⁴ T
- e = 1.5 × 10⁻¹⁹ C
- mₑ = 9.1 × 10⁻³¹ kg
ν = (1.5 × 10⁻¹⁹ C × 6.5 × 10⁻⁴ T) / (2 * π * 9.1 × 10⁻³¹ kg)
ν = (9.75 × 10⁻²³) / (57.18 × 10⁻³¹) Hz
ν ≈ 0.1705 × 10⁸ Hz
ν ≈ 17.05 × 10⁶ Hz = 17.05 MHz
Dependence on Speed
No, the frequency of revolution does not depend on the speed of the electron.
Explanation:
As derived above, the formula for the frequency of revolution is:
ν = (e * B) / (2 * π * mₑ)
This expression contains the charge of the electron (e), the mass of the electron (mₑ), and the magnetic field strength (B). It does not include the speed (v) or the radius (r) of the orbit. Therefore, for a given charged particle in a specific magnetic field, the frequency of revolution is constant, regardless of its speed. If the speed increases, the radius of the orbit increases proportionally, but the time taken for one revolution (and thus the frequency) remains the same.
Q13EXERCISES
(a) 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. (b) 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.)
Solution
(a) Calculation of Counter-Torque
To prevent the coil from turning, a counter-torque must be applied that is equal in magnitude and opposite in direction to the magnetic torque (τ) acting on the coil.
The magnitude of the magnetic torque is given by:
τ = N * I * A * B * sin(θ)
Where:
- N = number of turns = 30
- I = current = 6.0 A
- A = area of the coil
- B = magnitude of the magnetic field = 1.0 T
- θ = angle between the normal to the coil and the magnetic field = 60°
First, we calculate the area (A) of the circular coil:
- Radius, R = 8.0 cm = 0.08 m
- Area, A = π * R² = π * (0.08 m)² = 0.0064π m²
Now, substitute the values into the torque formula:
τ = 30 × 6.0 A × (0.0064π m²) × 1.0 T × sin(60°)
We know sin(60°) = √3 / 2 ≈ 0.866
τ = 180 × 0.0064π × 1.0 × (√3 / 2) N m
τ = 90 × 0.0064π × √3 N m
τ = 0.576π × √3 N m
τ ≈ 0.576 × 3.14 × 1.732 N m
τ ≈ 3.13 N m
The magnitude of the counter-torque required is 3.13 N m.
(b) Change for an Irregular Shape
No, the answer would not change.
Explanation:
The torque on a planar current loop in a uniform magnetic field depends on its magnetic moment (m), the magnetic field (B), and the angle (θ) between them (τ = mBsinθ). The magnetic moment is defined as m = N * I * A.
This formula for magnetic moment depends on the number of turns (N), the current (I), and the area enclosed by the loop (A). It does not depend on the specific shape of the loop. Since all other particulars, including the area, are unaltered, the magnetic moment of the irregular coil would be the same as that of the circular coil. Consequently, the torque experienced by the coil, and the required counter-torque, would remain the same.