Q 1 :    

AB is a part of an electrical circuit (see figure). The potential difference ''VA-VB'', at the instant when current i=2A and is increasing at a rate of 1 amp/second is   [2025]

[IMAGE 250]

  • 9 volt

     

  • 10 volt

     

  • 5 volt

     

  • 6 volt

     

(2)

[IMAGE 251]

From the diagram (Using KVL) 

       VA-Ldidt-5-2i=VB

Using, i=2A,  didt=1 A/sec and L=1 H, we get

VA-1-5-4=VB    VA-VB=10 V



Q 2 :    

Let us consider two solenoids A and B, made from same magnetic material of relative permeability μr and equal area of cross-section. Length of A is twice that of B and the number of turns per unit length in A is half that of B. The ratio of self inductances of the two solenoids, LA:LB is             [2024]

  • 1 : 2

     

  • 2 : 1

     

  • 8 : 1

     

  • 1 : 8

     

(1)

Self inductance of solenoid,

L=μ0N2Al=μrμ0n2lA

Where, n is number of turns per unit length,

A=Area and, l=length

Given: lA=2lB,  nA=12nB

   LALB=μrμ0lAAAnA2μrμ0lBABnB2=2lB·nB24nB2lB=12                      (AA=AB)



Q 3 :    

The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q0e-Rt/2L, where Q0 is the charge at t=0s. The time at which charge amplitude decreases to 0.50 Q0 is nearly 

[Given that R=1.5Ω,L=12 mH,ln(2)=0.693]              [2024]

  • 19.01 ms

     

  • 11.09 ms

     

  • 19.01 s

     

  • 11.09 s

     

(2)

Given, R=1.5Ω

L=12 mH;  ln 2=0.693

At t=0,  Q=Q0e-Rt/2L                             ...(i)

At t=t1,  Q=0.50 Q0                                  ...(ii)

 0.50 Q=Q0e-Rt1/2L

12=e-Rt1/2L  or  2=eRt1/2L

Taking log of both sides, ln2=Rt12L  or  0.693=1.5t24mH

   t1=11.09 ms



Q 4 :    

The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is      [2023]

  • 8 mJ

     

  • 8 μJ

     

  • 4 μJ

     

  • 4 mJ

     

(2)

As, U=12LI2=12×4×10-6×2×2;  U=8μJ

 



Q 5 :    

Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1R2, the mutual inductance M between them will be directly proportional to                 [2021]
 

  • R22R1

     

  • R1R2

     

  • R2R1

     

  • R12R2

     

(1)

Let a current I1 flows through the outer circular coil of radius R1.

The magnitude field at the centre of the coil is B1=μ0I12R1

As the inner coil of radius R2 is placed coaxially, therefore, B1 may be taken constant over its cross-sectional area.
Hence, flux associated with the inner coil is

ϕ2=B1πR22=μ0I12R1πR22 As M=ϕ2I1=μ0πR222R1MR22R1



Q 6 :    

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance           [2018]
 

  • 0.138 H

     

  • 138.88 H

     

  • 1.389 H

     

  • 13.89 H

     

(4)

Magnetic potential energy stored in an inductor is given by

U=12LI225×10-3=12×L×(60×10-3)2

L=25×2×106×10-33600=13.89H