JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 24)
A circular coil having N turns and radius r
carries a current I. It is held in the XZ plane in
a magnetic field B$${\mathop i\limits^ \wedge }$$ . The torque on the coil due
to the magnetic field is :
$${{B{r^2}I} \over {\pi N}}$$
B$$\pi $$r2IN
Zero
$${{B\pi{r^2}I} \over { N}}$$
Explanation
According to the question, the situation can be drawn as
_8th_April_Morning_Slot_en_24_1.png)
Let the current I is flowing in anti-clockwise direction, then the magnetic moment of the coil is
m = NIA
where, N = number of turns in coil
and A = area of each coil = $$\pi$$r2.
Its direction is perpendicular to the area of coil and is along Y-axis.
Then, torque on the current coil is
$$\tau = m \times B = mB\sin 90^\circ = NIAB$$
$$ = NI\pi {r^2}B(N - m)$$
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