Question

A coil of 100 turn and area 5 cm^2 is placed in a magnetic field of 0.2 t. the normal to the plane of the coil makes an angle of 60 with the direction of magnetic field. the magnetic flux linked with the coil is

Answer 1

Magnetic Flux =
$ba \cos( \alpha )$
where b = magnetic field
a = area

Magnetic fulx = 100×(0.2×5×10^-4 × (1/2))
= 0.0025wb

Answer 2

Explanation :

It is given that,

Number of turns of a coil, n = 100

Area of the coil, $A=5\ cm^2=0.0005\ m^2$

Magnetic field, B = 0.2 T

Angle between magnetic field and the normal to the plane is $\theta=60^0$

The magnetic flux is given by :

$\phi =nBA\ cos\theta$

n is the number of turns in the coil.

$\phi=100\times 0.2\ T\times 0.0005\ m^2\times\ cos(60)$

$\phi=0.005\ Wb$

Hence, this is the required solution.