Question

Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at t = 0. Find the emf induced in the loop at (a) t = 2 s, (b) t = 10 s, (c) t = 22 s and (d) t = 30 s.
Figure

Answer 1

Electromotive Force = $\frac {\phi}{\Delta}$

Explanation:

Given:

Initial velocity, u = 1 $\frac {cm}{s}$

Magnetic field, B = 0.6 T

(a) At t = 2 s:

Distance moved by the coil = $2 \times 1 \frac {cm}{s} = 2 cm = 2 \times 10^-^2 m$

Area under the magnetic field at t = 2s, A = $2 \times 5 \times 10^-^4 m^2$

Initial magnetic flux = 0

Final magnetic flux = BA = $0.6 \times (10 \times 10^-^4) Tm^2$

Change in the magnetic flux, $\Delta \phi = 0.6 \times (10 \times 10^-^4) − 0$

Now, induced emf in the coil is

e = $\frac {\phi}{\Delta}$

= $\frac {0.6 \times (10-0) \times 10^-^4}{2}$  

= $3 \times 10^-^4 V$

(b) At t = 10 s:

Distance moved by the coil = $10 \times 1$ = 10 cm

At this time square loop is completely inside the magnetic field, so there is no change in the flux linked with the coil with time.

Therefore, induced emf in the coil at this time is zero.

(c) At t = 22 s:

Distance moved = $22 \times 1$ = 22 cm

At this time loop is moving out of the field.

Initial magnetic flux = $0.6 \times (5 \times 5 \times 10^-^4)$ T-m

At this time 2 cm part of the loop is out of the field.  

Therefore, final magnetic flux = $0.6 \times (3 \times 5 \times 10^-^4)$ T-m

Change in the magnetic flux, $\Delta \phi = 0.6 \times (3 \times 5 \times 10^-^4) - 0.6 \times (5 \times 5 \times 10^-^4)$

=$- 0.6 \times 10^-^4 T-m^2$

e = $\frac {\phi}{\Delta}$

= $\frac {- 6 \times 10^-^4}{2}$

= $- 3 \times 10^-^4 V$

(d) At t = 30 s:

At this time loop is completely out of the field, so there is no change in the flux linked with the coil with time.

Therefore, induced emf in the coil at this time is zero.