 # 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=4t2 is switched on at time t=0 . The coefficient of friction between the ring and the table, if the ring starts rotating at t=2s , is:

Let us consider the coefficient of friction to be μ .

The ring is placed on a horizontal surface, while the magnetic field is vertical. So, the magnetic field is normal to the plane of the ring and hence parallel to its area vector. So the magnetic flux through the ring is given by

φ=BAcos0 …………………………….(1)

According to the question, the radius of the ring is R . So the area is given by

A = πR2

Also, the magnetic field is given as B=4t2. Therefore, from (1) the magnetic flux through the ring is

φ = (4t2)(πR2)cos0

⇒ φ =4πR2t2 …………………………….(2)

As can be seen above the flux through the ring is a function of time. So it varies with the time. We know that the changing magnetic flux through a surface generates an emf around its circumference. From Faraday’s law, the magnitude of this emf is given by

e = dφ/dt

From (2)

e = d(4πR2t2)/dt

⇒ e = 8πR2t …………………………….(3)

Now, we know that the electric field is related to the potential difference by

E = V/L

So the electric field along the circumference of the ring is

E=e/2πR

From (3)

E=8πR2t/2πR

⇒E=4Rt …………………………….(4)

Now, the ring has a uniform distribution of the charge q over its circumference. So the force on the charges is given by

F=qE

From (4)

F=4qRt …………………………….(5)

The electric force acting on a positive charge is known to act in the direction of the electric field. The electric field is present around the ring’s perimeter. As a result, the electric force operates tangentially throughout the ring’s whole circumference. The electric forces create couples that produce net torque around the ring’s centre, as can be shown. The source of this torque is

τ=FR

From (5)

τ=4qR2t …………………………….(6)

But there is frictional force also present between the horizontal surface and the ring which will produce a counter torque, given by

τ′ = fR …………………………….(7)

From (6) the electric torque increases with time. The frictional torque will keep the net torque zero until the external torque reaches the maximum value of the frictional torque. We know that the maximum value of the frictional force is given by

f = μN

As the ring is placed on a horizontal surface, so N=mg . Therefore

f = μmg

Substituting this in (7) we get

τ′ = μmgR …………………………….(8)

When the ring just starts rotating, the electric torque will be equal to the frictional torque, that is,

τ = τ′

From (6) and (8) we have

4qR2t = μmgR

⇒ μ = 4qR2t/mgR

On simplifying we get

μ = 4qRt/mg

According to the question, the time t = 2s . Substituting this above, we finally get

μ = 8qR/mg (1) (0)