Q99 / 180 · AIIMS 2007 · Physics

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

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

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

Hence, the correct answer is option: 'C'.

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