A current of `7*0A` is flowing in a plane circular coil of radius `1*0cm` having 100 turns. The coil is placed in a uniform magnetic field of `0*2Wb//m^2`. If the coil is free to rotate, what orientation would correspond to its (i) stable equilibrium and (ii) unstable equilibrium? Calculate potential energy of the coil in the two cases.

A current of `7*0A` is flowing in a plane circular coil of radius `1*0

1.	A current of `70A` is flowing in a plane circular coil of radius `10cm` having 100 turns. The coil is placed in a uniform magnetic field of `0*2Wb//m^2`. If the coil is free to rotate, what orientation would correspond to its (i) stable equilibrium and (ii) unstable equilibrium? Calculate potential energy of the coil in the two cases.
Answer» Here, `I=70A, r=10cm=10^-2m`, `N=100, B=02Wb//m^2` `M=NIA=Nipir^2` `=100xx70xx22/7(10^-2)^2=022Am^2` (i) Stable equilibrium corresponds to `theta=0^@` i.e. `vecM` parallel to `vecB`. `u_(min)=-MBcos 0^@=022xx02xx1` `=-0044J` (ii) unstable equilibrium corresponds to `vecM` antiparallel to `vecB`. P.E. is maximum `u_(max.)=-MB cos 180^@=-022xx02xx(-1)` `=0*044J`

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