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NCERT Class 12 Physics Chapter 4 Moving Charges And Magnetism
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Moving Charges And Magnetism
Chapter: 4
| Part – I |
EXERCISE
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?
Ans: Here,
n = 100,
r = 8 cm = 0.08 m
I = 0.40 A
B = μ0 × N × I / 2 × R
= (10-7 V 2 × 3 .14 V 0.4 × 100) / 8 × 10-2
= 3.1 × 10-5.
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?
Ans: The wire is long and it is considered as an infinite length wire, the magnetic field,
B = μ0 × I / 2𝜋r
μ = 4𝜋 × 10-7
I = 35 A
r = 20 cm or 0.20 m
B = 4 × 10-7 × 35 / 2𝜋 × 0.20
= 14 × 10-7 × 35 / 2 × 0.20
= 14 × 10-6 / 0.40
= 35 × 10-6
T = 35μ T.
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.
Ans: B = μ0 × I / 2𝜋r
μ = 4𝜋 × 10-7
I = 50 A
r = 2.5 m
B = 4𝜋 × 10-7 × 50 / 2𝜋 × 2.5
= 200 × 10-7 / 5
= 40 × 10-7 T = 40μ T.
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?
Ans: B = μ0 × I / 2𝜋r
μ = 4𝜋 × 10-7
I = 90 A
r = 1.5 m
B = 4𝜋 × 10-7 × 90 / 2𝜋 × 1.5
= 360 × 10-7 / 3
= 120 × 10-7 T
= 1.2 × 10-5 T
= 12 μT.
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 30º with the direction of a uniform magnetic field of 0.15 T?
Ans: Given:
F/L = I × B sin (θ)
I = 8 A.
θ = 30º
Here,
F/L = 8A × 0.15 T × sin 30º
(= Sin (30º) = ½ = 0.5)
= 8 × 0.15 × 0.5
= 8 × 0.075
= 0.6 N/m.
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?
Ans: Here:
I = 10 A
B = 0.27 T
L = 3.0 cm or 0.03m
θ = 90º
Given:
F = I × B × L × sin (θ)
= 10 A × 0.27 T × 0.03 m
= 0.081 N.
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.
Ans: Given:
μ = 4𝜋 × 10-7
IA and IB = 8.0 A and 5.0 B
d = 4.0 cm or 0.04 m.
Here,
F = μ0 × IA × IB / 2𝜋d
= 4𝜋 × 10-7 × 8.0 × 5.0 / 2 × 0.04
= 4 × 10-7 × 40.0 / 0.08
= 160 × 10-7 / 0.08
= 2 × 10-5 N/m
Calculate the total force (10 cm = 0.1m):
F = F × L
F = 2 × 10-5 × 0.1
= 2 × 10-6 N.
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.
Ans: Given:
μ0 = 4𝜋 × 10-7
I = 80 cm = 0.8 m
Total number of N = 5 × 400 = 2000 turns.
= n = N/L = 2000 / 0.8 m
= 2500 turns/m
Here:
B = μ0 × nl
= (4𝜋 × 10-7) × 2500 turns/m × 8.0
= 2.51 × 10-2 T.
9. A square coil of 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 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Ans: τ = n × l × A × B × sin θ
Slide of the square coil = 10 cm = 0.10 m
n = 20
l = 12 A
B = 0.80 T
Angle θ = 30º
Here,
A = side2 = (0.10)2 = 0.01 m2
τ = n × l × A × B × sin θ
= 20 × 12 × 0.01m2 × 0.80 × sin 30º
= 20 × 12 × 0.01 × 0.80 × 0.5
= 0.96 Nm.
10. Two moving coil metres, 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 metres). Determine the ratio of:
(a) Current sensitivity.
Ans: M1 = N1 × A1 × B1/ k
M2 = N2 × A2 × B2 /k
= Current sensitivity of M2 / M1
= (N2 × A2 × B2) / (N1 × A1 × B1)
= (42 × 1.8 × 10-3 × 0.50) / (30 × 3.6 × 10-3 × 0.25)
= 37.8 / 27
= 1.4
(b) Voltage sensitivity of M2 and M1.
Ans: M1 = N1 × A1 × B1 / kR1
M2 = N2 × A2 × B2 / kR2
= Current sensitivity of M2 / Current sensitivity of M1
= (N2 × A2 × B2 / R2) / (N1 × A1 × B1/R1)
= (N2 × A2 × B2 × R1) / (N1 × A1 × B1 × R2)
= (42 × 1.8 × 10-3 × 0.50 × 10) / (30 × 3.6 × 10-3 × 0.25 × 14)
= 378 / 378
= 1.
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, me = 9.1×10–31 kg)
Ans: Given:
B = 6.5 G = 6.5 × 10-4 T
v = 4.8 × 106 m/s
e = 1.6 × 10-9 C
me = 9.1 × 10-31 kg
Here:
F centripetal = me × v2 / r
e × v × B = me × v2 / r
Rearranging the formula to solve r:
r = me × v / e × B
= (9.1 × 10-31 kg × 4.8 × 106 m/s) / (1.6 × 10-19 C × 6.5 × 10-4 T.)
= 9.1 × 4.8 × 10-31+6 / (1.6 × 6.5 × 10-19-4)
= 4.2 × 10-2
= 4.2 cm.
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.
Ans: Given:
B = 6.5 G = 6.5 × 10-4 T
e = 1.6 × 10-9 C
me = 9.1 × 10-31 kg
f = eB/ 2𝜋me
= (1.6 × 10-19 × 6.5 × 10-4) / (2 × 3.14 × 9.1 × 10-31)
= 18.18 × 106 Hz
= 18 MHz.
13. (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.
Ans: τ = n × l × A × B × sinθ
n = 30
r = 8.0 cm = 0.08 m
l = 6.0 A
B = 1.0 T
Angle θ = 60°
Here,
A = 𝜋r2 = 𝜋 × (0.08)2 = 𝜋 × 0.0064
= 0.0201 m2
τ = n × l × A × B × sinθ
(Angle θ = 60° = √3/2)
= 30 × 6.0 × 0.0201 × √3/2
= 30 × 6.0 × 0.0201 × 0.866
= 30 × 0.1076
= 3.23 Nm.
(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).
Ans: No, the answer would not change. This is because the magnetic force acting on each infinitesimal element of the coil depends only on the current flowing through that element and the magnetic field at its location.
