Question

A line charge l per unit length is lodged uniformly onto the rim of awheel of mass M and radius R. The wheel has light non-conductingspokes and is free to rotate without friction about its axis (Fig. 6.22).A uniform magnetic field extends over a circular region within therim. It is given by,B = – B0 k (r £ a; a < R)= 0 (otherwise)What is the angular velocity of the wheel after the field is suddenly switched off?

Answer 1

The mass of the wheel is given as M, radius of the wheel is r.

The magnetic field is given as, B=− B 0 k ^ { r≤a≤R }.

Faraday’s law of electromagnetic induction gives the relation between induced emf and flux as,

ε=− dϕ dt

The relation between electric field and change of flux can be written as,

∫ E → ⋅ dl → =− dϕ dt (1)

The flux linked is calculated as,

dϕ dt =− d(π a 2 B) dt [ ∵ϕ=B×Area ] =−π a 2 dB dt

Substitute this value in equation (1) and simplify.

E ∫ dl → =−π a 2 dB dt E( 2πa )=−π a 2 dB dt E=− a 2 dB dt

Line charge per unit length is,

λ= Total charge Total length = Q 2πr

The force due to magnetic field is,

F M =QE =2πaλ( − a 2 dB dt ) =−π a 2 λ dB dt

The centripetal force on the particle due to rotation of the wheel is,

F=m⋅a =m dv dt =m d( Rω ) dt =mR dω dt

The magnetic force on the wheel is balanced by the centripetal force on the wheel.

mR dω dt =−π a 2 λ dB dt mRdω=−π a 2 λdB

Integrate the above equation and simplify.

∫ mR dω= ∫ −π a 2 λ dB ω=− π a 2 λB mR

For { r≤a≤R }, the angular velocity of the wheel in vector form is,

ω=− π a 2 λB mR k ^