With the decrease of current in the primary coil from 2 amperes to zero value in (0.01 ~s) the emf generated in the secondary coil is 1000 volts. The mutual inductance of the two coils is

Correct answer: C. (5.00 ~H) Induced (emf =(-M d I_P)/(d t)) Where M is the mutual inductance of two coils and ((d I_P)/(d t)) is the rate of change of current in the primary. aligned & 1000 V= -M(0 A-2 A) 0.01 s & or M= (1000 V)(0.01 s) (2 A) =5 H . aligned )

AIIMS 2007

Exam: AIIMS
Year: 2007
Subject: Physics
Chapter: Electromagnetic Induction

AIIMS 2007 · Physics · Electromagnetic Induction

With the decrease of current in the primary coil from 2 amperes to...

With the decrease of current in the primary coil from 2 amperes to zero value in (0.01 \mathrm{~s}) the emf generated in the secondary coil is 1000 volts. The mutual inductance of the two coils is

Options

A(1.25 \mathrm{~H})
B(2.50 \mathrm{~H})
C(5.00 \mathrm{~H})
D(10.00 \mathrm{~H})

Correct answer

C. (5.00 \mathrm{~H})

Step-by-step solution

Induced (\operatorname{emf} \varepsilon=\frac{-M d I_P}{d t}) Where

M

is the mutual inductance of two coils and (\frac{d I_P}{d t}) is the rate of change of current in the primary.

\begin{aligned} & \therefore \quad 1000 \mathrm{ V}=\frac{-M(0 \mathrm{ A}-2 \mathrm{ A})}{0.01 \mathrm{ s}} \\ & \text{ or } M=\frac{(1000 \mathrm{ V})(0.01 \mathrm{ s})}{(2 \mathrm{ A})}=5 \mathrm{H} . \end{aligned}

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