A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass m and radius r and it is in a uniform vertical magnetic field B0, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity g, on two conducting supports at P and Q. When a current I is passed through the loop, the loop turns about the line PQ by an angle given by [figure]
- = rIB0/(mg)
- = 2 rIB0/(mg)
- = rIB0/(2mg)
- = mg/( rIB0)
Answer
(1) = rIB0/(mg)
Solution
Magnetic moment = I r2, normal to the loop (horizontal when the loop hangs vertically).
After turning by about PQ, the angle between and vertical B0 is (90 - ): magnetic torque = I r2B0 .
Gravity acts at the centre, at distance r below PQ: torque = mg r .
Equilibrium: I r2B0 = mgr = rIB0mg
Answer: (A)