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 an - Physics MCQ AIEEE 2012 2026 | ExamTest.live | ExamTest.live

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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:

A $1$
B $2$
C $3$
D $4$

Solution & Step-by-step Explanation

Consider a ring of radius r and thickness d r .Surface charge density σ = \frac{Q}{π R ^{2}} .Charge on ring d q = σ (2 π r d r) = \frac{Q}{π R ^{2}} 2 π r d r = \frac{2 Q r d r}{R ^{2}} .Equivalent current d i = \frac{d q}{T} = \frac{d q ω}{2 π} = \frac{2 Q r d r ω}{2 π R ^{2}} = \frac{Q ω r d r}{π R ^{2}} .Magnetic field at center due to ring d B = \frac{μ _{0} d i}{2 r} = \frac{μ _{0} Q ω r d r}{2 π R ^{2} r} = \frac{μ _{0} Q ω}{2 π R ^{2}} d r .Integrating from 0 to R : B = \int_{0}^{R} \frac{μ _{0} Q ω}{2 π R ^{2}} d r = \frac{μ _{0} Q ω}{2 π R ^{2}} [r]_{0}^{R} = \frac{μ _{0} Q ω R}{2 π R ^{2}} = \frac{μ _{0} Q ω}{2 π R} .If Q and ω are constant, B \propto \frac{1}{R} .The graph is a rectangular hyperbola.

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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:

A

1

B

2

C

3

D

4

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