This set of Class 12 Physics Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Solenoid and Toroid”.
1. Pick out the expression for magnetic field inside a toroid from the following.
a) B = μ0 NI
b) B = 2μ0 NI
c) B = \(\frac {\mu_o}{NI}\)
d) B = \(\frac {\mu_o N}{I}\)
View Answer
Explanation: Ampere’s circuital law:
∮ B.dl = ∮ B.dlcos0 = B ∮ dl = B.2πr …………………..1
But, for a toroid:
∮ B.dl = μ0 × total current threading the toroid
∮ B.dl = μ0 × total number of turns in the toroid × I
∮ B.dl = μ0 N2πrI ……………………………2
From 1 and 2
B.2πr = μ0 N2πrI
Therefore, B = μ0NI
2. A solenoid of 0.5 m length with 100 turns carries a current of 5 A. A coil of 20 turns and of radius 0.02m carries a current of 0.6 A. What is the torque required to hold the coil with its axis at right angle to that of solenoid in the middle point of it?
a) 1.60 N m
b) 15.89 × 10-5 N m
c) 1.893 × 10-5 N m
d) 1.893 × 10-8 N m
View Answer
Explanation: Given: μ0 = 4 × 3.14 × 10-7; number of turns of solenoid = 100; Current passing through the solenoid = 5 A; Number of turns of coil = 20; Current passing through the coil = 0.6A; Length of the solenoid = 0.5 m; Radius of coil = 0.02 m
Magnetic field of solenoid (B) = μ0nI = μ0 × \(\frac {100}{0.5}\) × 5 = 4 × 3.14 × 10-7 × \(\frac {100}{0.5}\) × 5
Magnetic moment of the coil (M) = I × A × N = 0.6 × (0.02)2 × 20
Torque (τ) = MBsin (90o)
τ = 0.6 × (0.02)2 × 20 × 4 × 3.14 × 10-7 × \(\frac {100}{0.5}\) × 5 [sin (90o) = 1]
τ = 1.893 × 10-5 N m
3. Two concentric coils of 20 turns each are situated in the same plane. Their radii are 60 cm and 80 cm and they carry currents 0.4 A and 0.5 A respectively in opposite directions. Calculate the magnetic field at the center.
a) \(\frac {5\mu_0}{4}\)
b) \(\frac {5\mu_0}{6}\)
c) \(\frac {5\mu_0}{8}\)
d) \(\frac {5\mu_0}{10}\)
View Answer
Explanation: Since the two coils are concentric and in the same plane, carrying currents in opposite directions, the total magnetic field at the center of the concentric coils is given by:
B = B1 – B2 = \(\frac {\mu_0}{4\pi } [ \frac {2\pi N_1 I_1}{r_1} ] \, – \, \frac {\mu_0}{4 \pi } [ \frac {2 \pi N_2 I_2}{r_2} ]\)
B = \(\frac {\mu_0}{2} [ \frac {N_1 I_1}{r_1} \, – \, \frac {N_2 I_2}{r_2} ]\)
B = \(\frac {\mu_0}{2} \big [ \)20 × \( \frac {0.4}{0.6} \) – 20 × \( \frac {0.5}{0.8} \big ] \)
B = \(\frac {\mu_0}{2} [ \frac {40}{3} \, – \, \frac {25}{2} ]\)
B = \(\frac {\mu_0}{2} [ \frac {80-75}{6} ] \)
B = \(\frac {5\mu_0}{6}\)
4. Magnetic field in toroid is stronger than that in solenoid.
a) True
b) False
View Answer
Explanation: Yes, in the magnetic field at the center in toroid is stronger than that in a solenoid. This is due to its ring structure. The magnetic field inside and outside the toroid is zero. Toroid does not have a uniform magnetic field inside it unlike a solenoid.
5. A long solenoid has 500 turns per cm and carries a current I. The magnetic field at its center is 7.54 × 10-2 Wb m-2. Another long solenoid has 300 turns per cm and it carries a current \(\frac {I}{3}\). What is the value of magnetic field at the center?
a) 1.50 × 10-3 Wb/m2
b) 1.50 × 10-5 Wb/m2
c) 1.05 × 10-3 Wb/m2
d) 1.50 × 10-2 Wb/m2
View Answer
Explanation: The magnetic field induction at the center of a long solenoid is B = μonI→B ∝ nI n = number of turns per unit length of solenoid. I = current passing through the solenoid.
\(\frac {B1}{B2} = \frac {n1}{n2} [ \frac {I1}{I2} ] \)
\(\frac {B1}{B2} = \frac {500}{300} \, \times \, \frac {I}{(\frac {I}{3})}\)
\(\frac {B1}{B2}\) = 5
B2 = \(\frac {B1}{5}\)
B2 = \(\frac {7.54 \, \times \, 10^{-2}}{5}\)
B2 = 0.01508 = 1.50 × 10-2Wb/m2
6. A solenoid has core of a material with relative permeability 200 and its windings carry a current of 2 A. The number of turns of the solenoid is 200 per meter. What is the magnetization of the material?
a) 6 × 104 A/m
b) 7 × 104 A/m
c) 8 × 104 A/m
d) 9 × 104 A/m
View Answer
Explanation: Given: number of turns (n) = 200 turns per meter; current (I) = 2 A; Relative permeability (µr) = 200
Magnetic intensity is given by: H = nI = 200 × 2 = 400 A/m
μr = 1 + γ → γ is the magnetic susceptibility of the material
γ = μr – 1
Magnetization → M = γ x H
M = (μr – 1) × H
M = (200 – 1) × 200
M = 199 × 200
M = 79600
M = 7.96 × 104 A/m ≈ 8 × 104 A/m
7. A rectangular coil, of sides 4 cm and 5 cm respectively, has 50 turns in it. It carries a current of 2 A, and is placed in a uniform magnetic field of 0.5 T in such a manner that its plane makes an angle of 60o with the field direction. Calculate the torque on the loop.
a) 5 × 10-4 N m
b) 5 × 10-2 N m
c) 5 × 10-3 N m
d) 5 × 10-5 N m
View Answer
Explanation: The magnitude of torque experienced by a current carrying coil kept in a uniform magnetic field is given by:
τ = MB sin (θ) = (NIA) × B sin (θ)
M → Magnetic moment of the coil
N → Number of turns
I → Current in coil
A → Area of the coil
θ → Angle between normal to the plane of the coil and the direction of the magnetic field
Given: N = 50; A = 4 × 5 = 20 cm2 = 20 × 10-4 m2; B = 0.5 T; I = 2 A
θ = 90o – 60o = 30o
τ = (NIA) × B sin (θ)
τ = 50 × 2 × 0.5 × 20 × 10-4 × sin (30o)
τ = 0.05
τ = 5 × 10-2 N m
Sanfoundry Global Education & Learning Series – Physics – Class 12.
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