Question

A conducting square loop having edges of length 2.0 cm is rotated through 180° about a diagonal in 0.20 s. A magnetic field B exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field.

Answer 1

Magnetic field intensity is 5T

Explanation:

Given:

Average induced emf in the loop $\epsilon= 20 mV = 2 \times 10^-^2 V$

Time is taken to rotate the loop $\Delta t = 0.2 s$

Edge length of square loop $= 2 cm = 0.02 m$

Area of square loop A = $0.02^2 = 4 \times 10^-^4$

We know that,

Average induced emf in time interval $\Delta t$ is given by

$\epsilon = -(\phi_2 - \phi_1)$ """(i)

Where,

$\phi_2\ and\ \phi_1$are flux across the cross section at time intervals $t_1\ and\ t_2$ respectively.

Magnetic flux (\phi) through the loop is given by the formula

$\phi =\vec{B}\vec{A}$

$\phi = BAcos\theta$

Where B = magnetic field intensity

A = area of cross-section

θ = angle between area vector and magnetic field

Initially, the angle between area vector and the magnetic field is 0°

Therefore, Initial flux through the coil is  

$\phi_1 = BAcos0 = BA$

When it is rotated by 180 flux passing through the coil is given by

$\phi_2 = BAcos180 = - BA$

Putting this values in eqn.(i) we get,

$\epsilon = - \frac{(\phi_2 - \phi_1)}{\Delta t} = \frac {- BA - BA}{\Delta t} = \frac {2BA}{\Delta t}$

Putting the values of \epsilon , \vec{B}\ and $\Delta t$ in the above eqn-

$2 \times 10^-^2 = 2 \times B \times 4 \times \frac {10^-^4}{0.2}$

$B = 20 \times \frac {10^-^3}{4 \times 10^-^3}=5T$

Therefore, magnitude of magnetic field intensity is 5T