What is magnetic force on the ring at the instant when speed of ring is v :- | TestHub

Passage / Comprehension

There is a thin aluminium doughnut of square cross-section A and radius r. It is released from rest and falls vertically under gravity through horizontal radial magnetic field B produced by north, N and south, S magnetic poles as shown in figure. The ring has mass m, radius r and resistivity $ρ$ . The centre of the ring falls down the vertical line of symmetry of the system.

What is magnetic force on the ring at the instant when speed of ring is v :-

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Solution:

The motional emf around the ring is $ε = \oint (v \times B) \cdot d ℓ = 2 π r B v$ , since $v \times B$ is along $d ℓ$ everywhere.

Ring resistance: $R = \frac{ρ ( 2 π r )}{A}$ , so the induced current is $I = \frac{2 π r B v}{R} = \frac{B v A}{ρ}$ .

Magnetic force: $F = B I (2 π r) = \frac{2 π A B ^{2} v r}{ρ}$ . By Lenz's law (Fleming's left-hand rule) it opposes the fall, i.e. it acts upward.

Physics - EMI/AC Question with Solution | TestHub

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