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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACMotional & Rotational EMFHard2 minQB

PhysicsHardsingle choice

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.

Heat dissipated per unit time after achieving terminal velocity of ring :

Options:

Answer:

Solution:

At terminal velocity the magnetic force balances the weight, so the gravitational power $m g v_{T}$ is entirely converted to heat.

Using $v_{T} = \frac{m g ρ}{2 π A B ^{2} r}$ : $P = m g v_{T} = \frac{m ^{2} g ^{2} ρ}{2 π A B ^{2} r}$ .

(Equivalently $P = I^{2} R$ with $I = \frac{B v _{T} A}{ρ}$ and $R = \frac{2 π r ρ}{A}$ gives the same result.)

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Motional & Rotational EMF

⏱ 2mℹ️ Source: QB

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