Magnetics Ques 84
- Consider $d \gg a$, and the loop is rotated about its diameter parallel to the wires by $30^{\circ}$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
(a) $\frac{\mu _0 I^{2} a^{2}}{d}$
(b) $\frac{\mu _0 I^{2} a^{2}}{2 d}$
(c) $\frac{\sqrt{3} \mu _0 I^{2} a^{2}}{d}$
(d) $\frac{\sqrt{3} \mu _0 I^{2} a^{2}}{2 d}$
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Answer:
Correct Answer: 84.(b)
Solution:
Formula:
- Magnetic field at mid-point of two wires $=2$ (magnetic field due to one wire)
$ =2 [\frac{\mu _0}{2 \pi} \frac{I}{d}]=\frac{\mu _0 I}{\pi d} \otimes $
Magnetic moment of loop $M=I A=I \pi a^{2}$
Torque on loop $=M B \sin 150^{\circ}$
$ =\frac{\mu _0 I^{2} a^{2}}{2 d} $
BETA
