Question

In figure (a) and figure (b) two air-cored solenoids
P and Q have been shown. They are placed near
each other. In figure (a), when Ip, the current in P,
changes at the rate of 5 As, an emf of 2 mV is
induced in Q. The current in P is then switched off,
and a current changing at 2 Asl is fed through
Q as shown in diagram. What emf will be induced
in P:-
Р
lenne
mom
teveel
Q
Р
(a)
(b)
(1) 8 x 104 V
(3) 5 x 10-3 v
(2) 2 x 108V
(4) 8 x 10-2 V

Answer 1

Answer:

It is a case of Mutual Inductance.

The mutual inductance M may be defined as it is numerically equal to the induced emf in the secondary coil, when the current in the primary coil is changing at the rate of one ampere per second.

e=M

dt

di

In first case, 2mv=M×5

In second case, x=M×2 ,where x is emf induced in P

By solving both, we get

x=0.8mV

Answered By

Answer 2

$8 \times 10 - b$

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