Question

A rectangular coil of wire with a dimension of 4 cm x 5 cm and 10 turns is located between the poles of a large magnet that produces a uniform magnetic field of 0.75 T. The surface of the coil which is originally parallel to the field is rotated in 0.10 s, so that its surface is perpendicular to the field. Calculate the average induced emf across the ends of coil as the coil rotates.

Answer 1

Answer:0.15 V

Explanation:

Given

Dimension of coil $4cm\times 5cm$

Area of coil $A=4\times 5=20\ cm^2$

Magnetic field $B=0.75\ T$

Time of rotation $t=0.1\ s$

No of turns $N=10$

Initial flux associated with the coil

$\phi_i=N(B\cdot A)$

$\phi_i=N(BA\cos \theta )$

where $\theta$=angle between magnetic field and area vector of coil

$\phi_i=N(BA\cos 90 )$

Finally when coil is perpendicular to the field

$\phi_f=N(B\cdot A)$

$\phi_i=N(BA\cos 0 )$

and induced emf is given by

$e=-\frac{d\phi }{dt}$

$e=-\frac{\phi_1-\phi_2}{t-0}$

$e=-\frac{(0-10\times 0.75\times 20\times 10^{-4})}{0.1}$

$e=0.15\ V$