JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 24)
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be $$\sqrt x $$ $$\times$$ 10$$-$$5 Nm. The value of x is .................
Answer
3
Explanation
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$$\overrightarrow \tau = \overrightarrow M \times \overrightarrow B = MB\sin 90^\circ $$
$$ = MB = {{i\sqrt 3 {l^2}} \over 4}B$$
$$ = \sqrt 3 \times {10^{ - 5}}$$ N $$-$$ m
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