1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A coil of radius R carries current i1. Another concentric coil of radius r(r<<R) carries current i2. Places of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. If the maximum kinetic energy of smaller coil when both are released is μ0i1i2πr2MRX(MR2+mr2). Find X?Masses of coils are M and m, respectively.

Open in App
Solution

## Magnetic induction at center due to current in larger coil is: B=μ0i2R......(i)Magnetic dipole moment of smaller coil: M=πr2i2......(ii)Initially M and B are at 90oSo, U1=−MBcos90o=0When the coils become coplanarU2=−MBcos0o=−MBDecrease in potential energy =U1−U2=MB=μ0i1i2πr22RThis decrease in potential energy converts into kinetic energy. Here the kinetic energies are maximum when the coils become coplanar. The two coils become coplanar. The two coils rotate due to their mutual interaction only and if one rotates clockwise, then the other coil rotates anticlockwise.Let ω1 and ω2 be the angular velocities of larger and smaller coils when they become coplanar. According to law of conservation of angular momentum: I1ω1=ω2I2And according to law of conservation of energy: 12I1ω12+12I2ω22=UFrom above two equations, we get:12I1(I2ω2I1)2+12I2ω22=U⇒12I2ω22(I2I1+1)=U⇒12I2ω22=UI1I1+I2So maximum kinetic energy of smaller coil is given by:⇒12I2ω22=(μ0i1i2πr22R)((1/2)MR2(1/2)MR2+(1/2)Mr2)=μ0i1i2πr2MR2(MR2+mr2)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Torque on a Magnetic Dipole
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program