COMPASSLearning

Talk to us Register Login

JEE Main · 2012 · 29 Apr 2012 · Offline · Magnetic Effects of Current

A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity

ω

. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure

(A)(see structure)
✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)

Answer: (A)

Solution

Model the spinning charged disc as nested rings. A ring of radius

x

and width

d x

holds charge

d q = σ 2 π x d x

with

σ = \frac{Q}{π R ^{2}}

, and turning at angular speed

ω

it carries a current

d i = \frac{d q ω}{2 π} = σ ω x d x

. Its field at the centre is

d B = \frac{μ _{0} d i}{2 x} = \frac{μ _{0} σ ω}{2} d x

, and integrating from

0

to

R

gives

B = \frac{μ _{0} σ ω R}{2} = \frac{μ _{0} Q ω}{2 π R}

. With the charge and the angular speed held fixed,

B \propto \frac{1}{R}

, so the graph is a rectangular hyperbola that falls as the radius grows.

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option

Practise more: all JEE Main PYQs on Magnetic Effects of Current →

More JEE Main questions on Magnetic Effects of Current