Question

Two coils X and Y are linked such that emf $E$ is induced in Y when the current in X is changing at the rate $I^{\prime }(=dI/dt)$. If a current $I_0$ is now made to flow through Y, the flux linked with X will be

• A. $E{I}_0{I}^{\prime }$
• B. $\frac{I_0{I}^{\prime }}{E}$
• C. $\left(\frac{E}{I_0}\right)I^{\prime }$
• D. $\left(\frac{E}{I^{\prime }}\right)I_0$

Answer 1

The correct option is D $\left(\frac{E}{I^{\prime }}\right)I_0$

Let $M$ be the mutual inductance,
We can write $E=M{I}^{\prime }$
We know that $E=-\frac{d{\Phi }_B}{dt}$
$\therefore {\Phi }_B=-\int Edt=-M\int I^{\prime }dt=-MI$ where$I$ is the current in the circuit.
Given $I=I_0$.
So ${\Phi }_B=-M{I}_0=-\left(\frac{E}{I^{\prime }}\right)I_0$.
Sign is indicative of the direction. So for magnitude only
${\Phi }_B=\left(\frac{E}{I^{\prime }}\right)I_0$