Question

Find the flux of earth's magnetic field of magnitude 5.00 3 1025 t through a square loop of area 20.0 cm2 (a) when the field is perpendicular to the plane of the loop, (b) when the field makes a 30.0° angle with the normal to the plane of the loop, and (c) when the field makes a 90.0° angle with the normal to the plane

Answer 1

Magnetic flux through a closed loop will be given by formula

`\phi = B.A = BAcos\theta`

here

B = magnetic field

A = area

`\theta`= angle between normal to plane and magnetic field

Now we will use here the above formula for all the given conditions

i) magnetic field is normal to the plane

magnetic flux = BAcos0

`\phi = 5.0031025 * 20* 10^(-4) * cos0`

`\phi = 0.01 Wb`

ii) magnetic field makes 30 degree with normal to the plane

magnetic flux = BAcos30

`\phi = 5.0031025 * 20* 10^(-4) * cos30`

`\phi = 8.66*10^(-3) Wb`

iii) magnetic field is perpendicular to the normal of the plane

magnetic flux = BAcos90

`\phi = 5.0031025 * 20* 10^(-4) * cos90`

`\phi = 0 Wb`