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PhysicsEMI/ACInduced Electric FieldHard2 minPYQ_2020

PhysicsHardnumerical

A non-conducting ring of mass $m = 4 kg$ and radius $R = 10 cm$ has a charge $Q = 2 C$ uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field $B = 4 t^{3} T$ is switched on at $t = 0$ . At $t = 5 s$ ring starts to rotate about the vertical axis through the centre. The coefficient of friction between the ring and the surface is found to be $\frac{k}{24}$ . Then the value of $k$ is

Answer:

18.00

Solution:

The induced electric field at the periphery of the ring is

$E = \frac{R}{2} \frac{d B}{d t}$

$\Rightarrow E = \frac{R}{2} (12 t^{2}) = 6 R t^{2}$

When the ring is about to slip,

$q E R = μ m g R$

$\Rightarrow 6 q R t^{2} = μ m g$

$\Rightarrow μ = \frac{6 q R t^{2}}{m g} = \frac{3}{4} = \frac{18}{24}$

$k = 18$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Induced Electric Field

⏱ 2mℹ️ Source: PYQ_2020

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