A conducting loop of radius $$\frac{10}{\sqrt{\pi}}$$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:

The induced emf in a loop is given by Faraday’s law: $$ \varepsilon = -\frac{d\phi}{dt} $$, where $$ \phi $$ is the magnetic flux and $$ \phi = B \cdot A $$. Since the loop is circular and perpendicular to the magnetic field, $$ A = \pi r^2 $$. The magnitude is $$ |\varepsilon| = \left| \frac{d\phi}{dt} \right| $$.

Given radius $$ r = \frac{10}{\sqrt{\pi}} $$ cm, converting to meters gives $$ r = \frac{10}{\sqrt{\pi}} \times 10^{-2} = \frac{0.1}{\sqrt{\pi}} $$ m, so the area is $$ A = \pi r^2 = \pi \left( \frac{0.1}{\sqrt{\pi}} \right)^2 = \pi \times \frac{0.01}{\pi} = 0.01 $$ m².

The magnetic field decreases linearly from 0.5 T to 0 T in 0.5 s at a steady rate, so $$ \frac{dB}{dt} = \frac{\Delta B}{\Delta t} = \frac{0 - 0.5}{0.5} = -1 \, \text{T/s} $$ and thus $$ \left| \frac{dB}{dt} \right| = 1 \, \text{T/s} $$.

Since $$ \phi = B \cdot A $$ and $$ A $$ is constant, $$ \frac{d\phi}{dt} = A \frac{dB}{dt} $$. The magnitude of induced emf is $$ |\varepsilon| = A \left| \frac{dB}{dt} \right| = 0.01 \times 1 = 0.01 \, \text{V} = 10 \, \text{mV} $$.

Because the rate of change of $$ B $$ is constant, the induced emf remains constant, so at $$ t = 0.25 $$ s it is still 10 mV.

The correct option is B. emf = 10 mV.