A conducting loop which is in the shape of a right-angled isosceles triangle of height $10 \mathrm{cm}$ is kept such that the $90°$ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in a counterclockwise direction and increased at a constant rate of $10\mathrm{ }\mathrm{A}\mathrm{ }{\mathrm{s}}^{-1}.$ The emf induced in the wire is

• A

  $\mathrm{}{\mu }_0\pi$
• B

  $2{\mu }_0\pi$
• C

  $\frac{2{\mu }_0}{\pi }$
• Correct answer
  D

  $\frac{{\mu }_0}{\pi }$

Solution

by direction of induced electric field, current in wire is in same direction of current along the hypotenuse.

Flux through triangle if wire have current i $= \underset{0}{\overset{0.1}{\int }}\left(\frac{{\mu }_{0}i}{2\pi x}\right) \left(2xdx\right)=\frac{{\mu }_{0}i}{10\pi }$

$\Rightarrow$ Mutual inductance $=\frac{{\mu }_0}{10\pi }$

Induced emf in wire $=\frac{{\mu }_0}{10\pi }\frac{di}{dt}=\frac{{\mu }_0}{10\pi }\times 10=\frac{{\mu }_0}{\pi }$

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