Question

A square coil of side $25cm$ having $1000$ turns is rotated with a uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant $t$, the emf induced in the coil is $e=200\sin \left(100\pi t\right)$. Magnetic induction is

• A. $0.50T$
• B. $0.02T$
• C. $10.3T$
• D. $0.1T$
• E. $0.01T$

Answer 1

The correct option is E

$0.01T$

Step 1: Given data

A square coil of side $25cmor0.25m$

$N=1000turns$

$e=200\sin 100\mathrm{\pi t}$

Area $$\begin{gathered} \left(A\right)=Side\times Side \\ A=0.25\times 0.25 \\ A=0.0625m^2 \end{gathered}$$

Step 2: Formula used:

The general formula of the conductor in a magnetic field is $E=vLB$

The negative sign indicates that

The general formula of the coil in a magnetic field is $E=BAN\omega \sin \theta =BAN\omega \sin \left(\omega t\right)$.

The maximum positive value of the e.m.f $\left(E_0\right)$ is when $\theta =\omega t=90°$ .

Where $\theta$ is angle between the plane of the armature and the magnetic field.

$Since,Sin\left({90}^0\right)=1$

So finally we get $E_0$=$BAN\omega$.

An emf is induced due to the motion of the conductor in a magnetic field which is given by $e=vlB$ where, $e$ is the induced emf, $l$ is the length of the conductor, $v$ is the velocity of the conductor.

The emf induced due to angular motion of the conductor is given by $BAN\omega =E_0$, where $A$ is the area, $B$ is the magnetic induction, $N$ is the number of turns, $E_0$ is the maximum value of emf and $\omega$ is the rotational speed.

Step 3: Calculate the magnetic induction

We have, $E_0=200\sin \left(100\pi t\right)$

$$\begin{gathered} \omega =100\pi \\ \therefore BAN\omega =E_0 \\ \therefore B=\frac{E_0}{AN\omega } \\ =\frac{200}{\left(0.25\times 0.25\right)\times 1000\times 100\pi } \end{gathered}$$

or $B=0.01T$

Hence, magnetic induction is $B=0.01T$.

Thus, the correct option is option E.