A planar loop of wire rotates in a uniform magnetic field. Initially, at $$t = 0$$, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

We recall Faraday’s law of electromagnetic induction, which states that the magnitude of induced emf in a loop is

$$\varepsilon = \left|\,\dfrac{d\Phi}{dt}\,\right|,$$

where $$\Phi$$ is the magnetic flux through the loop.

If a loop of area $$A$$ is placed in a uniform magnetic field of magnitude $$B$$ and the angle between the field and the normal (area vector) to the plane of the loop is $$\theta$$, then the flux is

$$\Phi = BA\cos\theta.$$

At $$t = 0$$ the plane of the loop is perpendicular to the magnetic field, so the normal is parallel to the field. Hence

$$\theta(0)=0^{\circ}\quad\Longrightarrow\quad\Phi(0)=BA\cos0^{\circ}=BA.$$

The loop rotates with a period $$T = 10\,\text{s}$$ about an axis lying in its own plane. That axis is perpendicular to the normal, so the normal vector performs a simple rotation about the field direction. The angular frequency is

$$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{10} = \dfrac{\pi}{5}\,\text{rad s}^{-1}.$$

Because at $$t=0$$ the normal is along the field, the instantaneous angle between the normal and the field after time $$t$$ is

$$\theta(t)=\omega t.$$

Therefore the flux as a function of time is

$$\Phi(t)=BA\cos(\omega t).$$

We now differentiate to find the induced emf:

$$\varepsilon(t)=\left|\,\dfrac{d}{dt}\bigl[BA\cos(\omega t)\bigr]\,\right| = \left|\,BA\bigl(-\omega\sin(\omega t)\bigr)\,\right| = BA\omega\,|\sin(\omega t)|.$$

The magnitude $$\varepsilon(t)$$ is maximum when

$$|\sin(\omega t)| = 1\quad\Longrightarrow\quad \sin(\omega t)=\pm1.$$

This occurs at

$$\omega t = \dfrac{\pi}{2},\,\dfrac{3\pi}{2},\,\dfrac{5\pi}{2},\ldots$$

Taking the first positive time, put $$\omega t = \dfrac{\pi}{2}$$:

$$t_{\text{max}}= \dfrac{\pi/2}{\omega}= \dfrac{\pi/2}{\pi/5}= \dfrac{5}{2}=2.5\,\text{s}.$$

The magnitude $$\varepsilon(t)$$ is minimum (in fact zero) when

$$|\sin(\omega t)| = 0\quad\Longrightarrow\quad \sin(\omega t)=0,$$

which happens for

$$\omega t = 0,\,\pi,\,2\pi,\ldots$$

Choosing the first non-zero positive time, set $$\omega t = \pi$$:

$$t_{\text{min}}= \dfrac{\pi}{\omega}= \dfrac{\pi}{\pi/5}=5\,\text{s}.$$

Thus, the induced emf attains its first maximum at $$2.5\ \text{s}$$ and its next minimum at $$5.0\ \text{s}$$.

Comparing with the given options, this corresponds to Option B.

Hence, the correct answer is Option B.