JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 19)
A square shaped coil of area $$70 \mathrm{~cm}^{2}$$ having 600 turns rotates in a magnetic field of $$0.4 ~\mathrm{wbm}^{-2}$$, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $$60^{\circ}$$ with the field, will be ____________ V. (Take $$\pi=\frac{22}{7}$$)
Answer
44
Explanation
Area $(\mathrm{A})=70 \mathrm{~cm}^2=70 \times 10^{-4} \mathrm{~m}^2$
$$ \mathrm{B}=0.4 \mathrm{~T} $$
$f=\frac{500 \text { revolution }}{60 \text { minute }}=\frac{500}{60} \frac{\text { rev. }}{\mathrm{sec} .}$
Induced emf in rotating coil is given by
$$ \begin{aligned} & e=N \omega B A \sin \theta \\\\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{60} \times 0.4 \times 70 \times 10^{-4} \sin 30^{\circ} \\\\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{6} \times 0.4 \times 70 \times 10^{-4} \times \frac{1}{2} \\\\ & =44 \text { Volt } \end{aligned} $$
$$ \mathrm{B}=0.4 \mathrm{~T} $$
$f=\frac{500 \text { revolution }}{60 \text { minute }}=\frac{500}{60} \frac{\text { rev. }}{\mathrm{sec} .}$
Induced emf in rotating coil is given by
$$ \begin{aligned} & e=N \omega B A \sin \theta \\\\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{60} \times 0.4 \times 70 \times 10^{-4} \sin 30^{\circ} \\\\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{6} \times 0.4 \times 70 \times 10^{-4} \times \frac{1}{2} \\\\ & =44 \text { Volt } \end{aligned} $$
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