A non-conducting ring of radius R and mass m having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying uniform magnetic field B = ( 5 qR mg ) t 2 is switched on at time t = 0 . The coefficient of friction between the ring and the table, if the ring starts rotating at t = 2 sec , is μ . Fill the value of 10 μ .

E = 2 R d t d B = 5 q m g t , driving torque τ = qER = 5 m g Rt Rotation starts when τ = μ m g R at t = 2 s: μ = 5 t = 0.4 , so 10 μ = 4

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PhysicsEMI/ACMiscellaneousMedium3 minQB

PhysicsMediuminteger

A non-conducting ring of radius R and mass m having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying uniform magnetic field $B = (\frac{mg}{5 qR}) t^{2}$ is switched on at time $t = 0$ . The coefficient of friction between the ring and the table, if the ring starts rotating at $t = 2 sec$ , is $μ$ . Fill the value of $10 μ$ .

diagram

Answer:

Solution:

$E = \frac{R}{2} \frac{d B}{d t} = \frac{m g t}{5 q}$

, driving torque

$τ = qER = \frac{m g Rt}{5}$

Rotation starts when

$τ = μ m g R$

at t = 2 s:

$μ = \frac{t}{5} = 0.4$

, so

$10 μ = 4$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Miscellaneous

⏱ 3mℹ️ Source: QB

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