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Question: Answered & Verified by Expert
A thin non conducting disc of mass M= 2 kg, charge Q=2×10-2 C and radius R=16 m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate system. A non-uniform, radial magnetic field B=B0 r^ exists in space, where B0=10 T and r^ is a unit vector in the radially outward direction. The disc is set in motion with an angular velocity ω=x×102 rad s-1, about an axis passing through its centre and perpendicular to its plane, as shown in the figure. At what value of x, the disc will lift off from the surface.

PhysicsMagnetic Effects of CurrentJEE Main
Solution:
1974 Upvotes Verified Answer
The correct answer is: 9
The force an any small part of the disc is in the vertically upward direction

dF=QπR22πr drωr B0

dF=2QωB0R2 r2 dr




F=23QωB0R=Mg 

ω=3Mg2QB0R=3×2×102×2×10-2×10×16=9×102 rad s-1

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