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  • A square loop of side ad resistance is placed in a magnet

A square loop of side \mathrm{1m} ad resistance \mathrm{1\Omega} is placed in a magnetic field of \mathrm{0.5T}. If the plane of the loop is perpendicular to the direction of the magnetic field, the magnetic flux through the loop is :

Option: 1

\mathrm{0.5weber}


Option: 2

\mathrm{1weber}


Option: 3

\mathrm{zero\;weber}


Option: 4

\mathrm{2weber}


Answers (1)

best_answer

\mathrm{B =0.5 T} \\

\mathrm{\phi =B A \cos 0^{\circ}} \\               \mathrm{(\because \bar{A}\| \bar{B}) } \\

\mathrm{\phi =B A=(0.5)(1)^{2}} \\

\mathrm{\phi =0.5 \text { weber }}

Hence correct option is 1

Posted by

Ritika Kankaria

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