Question

A circular loop of area 0.05 m^2 rotates in a uniform magnetic field of 0.2 T. If the loop rotates about its diameter which is perpendicular to the magnetic field,find flux linked with loop when its plane is inclined 60 degree to the field
A. 0.01wb
B. 0 Wb
C. 8.66x10-3Wb
D. 0.86wb
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Answer 1

Given:

A circular loop of area 0.05 m² rotates in a uniform magnetic field of 0.2 T. Plane is inclined 60° to the field.

To find:

Net magnetic flux ?

Calculation:

• Since the plane is inclined at 60° with the field, the area vector will be (90-60) = 30° with the field.

Now, magnetic flux is :

$\phi = B \times A \times \cos( \theta)$

$\implies \: \phi = 0.2 \times 0.05 \times \cos( {30}^{ \circ} )$

$\implies \: \phi = 0.2 \times 0.05 \times \dfrac{ \sqrt{3} }{2}$

$\implies \: \phi = 0.086 \: Wb$

$\implies \: \phi = 8.6 \times {10}^{ - 3} \: Wb$

OPTION C) is correct ✔️