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  • A square loop of side length 'a' is placed in a uniform magnetic field 'B' perpendicular to the plane of the loop. If the loop is rotated by the angle '<img alt="\theta" sr

A square loop of side length 'a' is placed in a uniform magnetic field 'B' perpendicular to the plane of the loop. If the loop is rotated by the angle '\theta' about an axis passing through its centre and parallel to one of its sides, what is the change in magnetic flux through the loop?

 

Option: 1

Zero


Option: 2

a^2 B \cos \theta


Option: 3

-a^2 B


Option: 4

a^2 B \cos \theta-a^2 B


Answers (1)

best_answer

The change in magnetic flux through the loop is given by \Delta \phi=B A \Delta(\cos \theta)

where A is the area of the loop. 

As the loop is rotated by the angle \theta, the change in \cos \theta is given by

\Delta(\cos \theta)=\cos \theta-\cos 0^{\circ}=(\cos \theta-1)

So, \begin{aligned} \Delta \phi & =B A \cdot(\cos \theta-1) \\ & =B \cdot a^2(\cos \theta-1) \\ & =B a^2 \cos \theta-B a^2 \end{aligned}

Posted by

Gaurav

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