A rectangular loop ( $\mathbf{area}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$) turns in a uniform magnetic field, B=0.20 T.When the angle between the field and the normal to the plane of the loop is $\frac{\mathbf{\tau \tau }}{\mathbf{2}}\mathbf{rad}$and increasing at0.60 rad/s, what emf is induced in the loop?

i) Area of the loop $A=0.15m^2$

ii) Magnetic field$B=0.20T$

iii) Angle$\theta =\frac{\pi }{2}rad=90°$

iv) Angular speed$\omega =0.60rad/s$

Use Faraday’s law to find the emf induced in the coil. Substituting the expression of the flux in the emf and substituting all the given values, find the required value of the induced emf.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

$\epsilon =-\frac{d\varphi }{dt}$

$\varphi =\overrightarrow{B}.\overrightarrow{A}$

Where,$\varphi$is magnetic flux, B is magnetic field, A is area, 𝜀 is emf.

Since, the loop is rotating continuously in the uniform magnetic field, the magnetic flux through the loop changes.

And this flux is given by,

$f=\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta$

The rate of change of the flux is nothing but the emf induced and by Faraday’s law, it is given by,

$\epsilon =-\frac{d\varphi }{dt}$

Where the negative sign shows that the magnetic field due to the currentopposes the change in the magnetic flux that induces the current.

Substituting the flux in Faraday’s law,

$\epsilon =-\frac{d}{dt}(BA\cos \theta )$

$\epsilon =BA\sin \theta \frac{d\theta }{dt}$

Since,$B=0.20T,A=0.15m^2,\omega =\frac{d\theta }{dt}=0.60\frac{rad}{s}$

Substituting all the values,

$\epsilon =0.20T\times 0.15m^2\times sin{90}^0\times 0.60\frac{rad}{s}$

$\epsilon =0.018V$

Hence, the emf induced in the loop is 0.018 V.

Therefore, using Faraday’s law, the emf induced in the coil can be determined.