A rectangular loop of wire is placed in the $X Y$-plane with its side of length $3 \mathrm{~cm}$ parallel to the $X$-axis and the side of length $4 \mathrm{~cm}$ parallel to the $Y$-axis. It is moving in the positive $X$-direction with the speed $10 \mathrm{~cm} / \mathrm{s}$. A magnetic field exists in the space with its direction parallel to the $Z$-axis. The field decreases by $2 \times 10^{-3} \mathrm{~T} / \mathrm{cm}$ along the positive $X$-axis and increases in time by $2 \times 10^{-2} \mathrm{~T} / \mathrm{s}$. The induced emf in the wire is

• A

  $-4.8 \times 10^{-5} \mathrm{~V}$
• B

  $4.8 \times 10^{-5} \mathrm{~V}$
• Correct answer
  C

  0
• D

  $3.6 \times 10^{-5} \mathrm{~V}$

Solution

Induced emf in wire = Rate of change of flux $$ \begin{aligned} & =\frac{d}{d t}(B A)=A\left\{\left(\frac{d}{d t} B\right)+v \frac{d}{d x} B\right\} \\ & =12 \times 10^{-4}\left(2 \times 10^{-2}+10 \times 10^{-2} \times 2 \times 10^{-3}\right) \\ & =12 \times 10^{-4}\left(2 \times 10^{-2}+2 \times 10^{-4}\right) \\ & =24 \times 10^{-4}\left(10^{-2}+10^{-4}\right)=24 \times 10^{-5} \mathrm{~V} \\ & =0.000024 \mathrm{~V} \approx 0 \mathrm{~V} \end{aligned} $$

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