Question

A circular coil has 60 turns and a radius of 1 cm. It is at the equator, where the earth's magnetic field is 0.7 G north. Find the magnetic flux through the coil when its plane is as follows.

(a) horizontal

______Wb

(b) vertical with its axis pointing north

______Wb

(c) vertical with its axis pointing east

______Wb

(d) vertical with its axis making an angle of 30° with north

______Wb

Answer 1

no of the coil N=60

radius of the coil r=1cm=0.01cm

magnetic field B=0.7Gauss=0.7*10^-4 T

magnetic flux=N*B*A*cos(theta)

here, theta is angle between B and plane of the area,

a)

if plane of the coil is horizontal

theta=90 degrees

magnetic flux=N*B*A*cos(theta)

=60*0.7*10^-4(pi*0.01^2)*cos(90)

=0 Wb

b)

if plane of the coil is vertical and its axis is pointing north

theta=0 degrees

magnetic flux=N*B*A*cos(theta)

=60*0.7*10^-4(pi*0.01^2)*cos(00)

=1.32*10^-6 Wb

c)

if plane of the coil is vertical and its axis is pointing east

theta=90 degrees

magnetic flux=N*B*A*cos(theta)

=60*0.7*10^-4(pi*0.01^2)*cos(90)

=0 Wb

d)

if plane of the coil is vertical and its axis is making 30 degrees with north

theta=30 degrees

magnetic flux=N*B*A*cos(theta)

=60*0.7*10^-4(pi*0.01^2)*cos(30)

=1.143*10^-6 Wb