Question

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

• A. $2.5secand5sec$
• B. $5secand7.5sec$
• C. $2.5secand7.5sec$
• D. $5secand10sec$

Answer 1

The correct option is A

$2.5secand5sec$

Step 1. Given Data:

Time period, $T=10s$

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

Let $\omega$ is frequency, $T$ is time period, $e$ is induced electromagnetic frequency $\left(emf\right)$ and ${\phi }_{flux}$ be magnetic flux.

Step 2. To calculate $\omega$:

We know that,

Frequency $\omega =2\mathrm{\pi }/$Time period $T$

$$\begin{gathered} \omega =\frac{2\pi }{T} \\ \Rightarrow \omega =\frac{2\pi }{10} \\ \Rightarrow \omega =\frac{\pi }{5} \end{gathered}$$

Step 3. Time $t$ at which magnitude of induced emf will be maximum and minimum,

We know that,

Magnetic flux ${\phi }_{flux}=\overrightarrow{B}·\overrightarrow{A}$

$$\begin{gathered} \Rightarrow {\phi }_{flux}=B·A\cos \theta \\ \Rightarrow {\phi }_{flux}=B·A\cos \omega t\dots \left(1\right) \end{gathered}$$

Induced $emf$, $\left|e\right|=\left|\frac{d{\phi }_{flux}}{dt}\right|$,

Differentiating equation $\left(1\right)$ with respect to $t$ we get $\left|e\right|,$

$$\begin{gathered} \left|e\right|=\left|\frac{d{\phi }_{flux}}{dt}\right| \\ \Rightarrow \left|e\right|=B·A\sin \omega t\dots \left(3\right) \end{gathered}$$

From equaton $\left(3\right)$ maximum $\left|e\right|$ will be at, $\omega t=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \omega =\frac{\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{t} \\ \Rightarrow t=\frac{\left({\displaystyle \frac{\pi }{2}}\right)}{\left(\frac{\pi }{5}\right)} \\ \Rightarrow t=2.5sec \end{gathered}$$

Time $t$ at which magnitude of induced emf will be maximum$2.5sec$.

From equaton $\left(3\right)$ minimum $\left|e\right|$ will be at, $\omega t=\mathrm{\pi }$

$$\begin{gathered} \omega =\frac{\mathrm{\pi }}{t} \\ \Rightarrow t=\frac{\mathrm{\pi }}{\left(\frac{\pi }{5}\right)} \\ \Rightarrow t=5sec \end{gathered}$$

Time $t$ at which magnitude of induced emf will be minimum$5sec$.

Hence, the correct option is (A).