Q 1 :

In a coil, the current changes from -2A to +2A in 0.2 s and induces an emf 0.1 V. The self-inductance of the coil is:               [2024]

  • 4 mH

     

  • 5 mH

     

  • 2.5 mH

     

  • 1 MH

     

(2)     

           (Emf)induced=-Ldidt

           In magnitude form,

           |Emfind|=|(-)Ldidt|

           0.1=(L)[+2-(-2)]0.2L=0.1×0.24=5mH

 



Q 2 :

The current in an inductor is given by I=(3t+8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self-inductance of the inductor ________ mH.                 [2024]



(4)      ||=Ldidt

          i=|3t+8|didt=3

          ε=12mV

          |ε|=L|dIdt|12=L×3

          L=4mH

 



Q 3 :

Two coils have mutual inductance 0.002H. The current changes in the first coil according to the relation i=i0sinωt, where i0=5A and ω=50π rad/s. The maximum value of emf in the second coil is πα V. The value of α is _____.                              [2024]



(2)     ϕ=Mi=Mi0sinωt

         EMF=-Mdidt=-0.002(i0ωcosωt)

         EMFmax=i0ω(0.002)=(5)(50π)(0.002)

         EMFmax=π2V



Q 4 :

Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is           [2024]

  • μ0πb22a

     

  • μ0πa22b

     

  • μ02π·b2a

     

  • μ02π·a2b

     

(2)

Magnetic flux through smaller loop, ϕ=Mi=BA

ϕ=(μ0i2b)·πa2=Mi

M=μ0πa22b



Q 5 :

A small square loop of wire of side L is placed inside a large square loop of wire of side L(L=l2). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is x×10-7H, where x= ____.                  [2024]



(128)

Flux linkage for inner loop.

ϕ=Bcenter·l2=4×μ0i4πL2(sin45+sin45)l2

ϕ=22μ0iπLl2

M=ϕi=22μ0l2πL=22μ0π

=224ππ×10-7=128×10-7 H

x=128



Q 6 :

Regarding self-inductance:

A. The self-inductance of the coil depends on its geometry.

B. Self-inductance does not depend on the permeability of the medium.

C. Self-induced e.m.f. opposes any change in the current in a circuit.

D. Self-inductance is electromagnetic analogue of mass in mechanics.

E. Work needs to be done against self-induced e.m.f. in establishing the current.

Choose the correct answer from the options given below:          [2025]

  • A, B, C, D only

     

  • A, C, D, E only

     

  • A, B, C, E only

     

  • B, C, D, E only

     

(2)

Self-inductance of coil

L=μ0μrN22πR



Q 7 :

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 and 2, respectively. If L1 = self-inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be          [2025]

  • ε1=L1dI1dt+M12dI2dt

     

  • ε1=L1dI1dtM12dI1dt

     

  • ε1=L1dI1dtM12dI2dt

     

  • ε1=L1dI2dtM12dI1dt

     

(3)

ϕ1=L1I1+M12I2

ε1=dϕ1dt=L1dI1dtM12dI2dt

Magnitude of induced emf due to self-inductance, =LdI1dt

Magnitude of induced emf due to mutual inductance, =MdI2dt



Q 8 :

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'R' is placed inside a large square loop of wire of side L (L>>R). The loops are coplanar and their centres coincide.                  [2023]

  • M=2μ0R2L

     

  • M=22μ0RL2

     

  • M=22μ0R2L

     

  • M=2μ0RL2

     

(3)

ϕ=Mi  also  ϕ=BA

ϕ=πR2(4μ04πi(L/2)2)

  M=22μ0R2L



Q 9 :

A 12 V battery connected to a coil of resistance 6Ω through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is                     [2023]

  • 5 mH

     

  • 12 mH

     

  • 8 mH

     

  • 10 mH

     

(4)

Induced emf=-LdIdt

  20=-L(0-2)10-3

  L=10 mH



Q 10 :

Two concentric circular coils with radii 1 cm and 1000 cm, and number of turns 10 and 200 respectively, are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _________ ×10-8 H. (Take, π2=10)                   [2023]



(4)

r1=1 cm,  N1=10

r2=1000 cm,  N2=200

ϕ1,2=MI2

N2B2·N1A1=MI2

  N1N2μ0I22r2·πr12=MI2

  M=10×200×4π×10-7×π×(0.01)22×10

  M=4×10-8