A non conducting ring(mass m,radious r,having charge Q) is placed on a rough horizontal surface(in a cylindrical region with tranverse magnetic field ).The field is increasing with time at the rate R and the coefficient of friction b/w the surface and the ring is μ.For ring to remain in equilibrium μ should be greater than
(1)Qr^2R^2/2mg
(2)QrR/2mg
(3)Qr^2R/2mg
(4)QrR^2/2mg
solver it and say how you have done?

To determine the condition for the non-conducting ring to remain in equilibrium while placed on a rough horizontal surface in a transverse magnetic field that is increasing with time, we need to analyze the forces and torques acting on the ring. Let's break this down step by step.

Understanding the Forces at Play

The ring has a mass \( m \), a radius \( r \), and carries a charge \( Q \). When the magnetic field \( B \) increases at a rate \( R \), it induces an electromotive force (emf) in the ring due to Faraday's law of electromagnetic induction. This induced emf generates a current \( I \) in the ring, which in turn produces a magnetic force.

Induced Current and Magnetic Force

According to Faraday's law, the induced emf \( \mathcal{E} \) in the ring can be expressed as:

• \(\mathcal{E} = -\frac{d\Phi_B}{dt}\)

Where \( \Phi_B \) is the magnetic flux through the ring. The magnetic flux can be calculated as:

• \(\Phi_B = B \cdot A\)

For a ring of radius \( r \), the area \( A \) is \( \pi r^2 \). If the magnetic field is increasing at a rate \( R \), we have:

• \(\mathcal{E} = -\pi r^2 \frac{dB}{dt} = -\pi r^2 R\)

The induced current \( I \) in the ring can be calculated using Ohm's law:

• \( I = \frac{\mathcal{E}}{R_{ring}} \)

Where \( R_{ring} \) is the resistance of the ring. The magnetic force \( F_B \) acting on the ring due to the magnetic field is given by:

• \( F_B = I \cdot B \cdot L \)

Here, \( L \) is the length of the wire, which for a ring is the circumference \( 2\pi r \). Thus, the magnetic force can be expressed as:

• \( F_B = I \cdot B \cdot 2\pi r \)

Frictional Force and Equilibrium Condition

For the ring to remain in equilibrium, the frictional force \( F_f \) must balance the magnetic force \( F_B \). The maximum static frictional force can be expressed as:

• \( F_f = \mu mg \)

Setting the magnetic force equal to the frictional force gives us the condition for equilibrium:

• \( \mu mg \geq F_B \)

Substituting Values

Now, substituting the expression for \( F_B \) into the equilibrium condition:

• \( \mu mg \geq I \cdot B \cdot 2\pi r \)

Substituting \( I \) and \( B \) into this equation, we can derive the necessary condition for \( \mu \):

• \( \mu mg \geq \left(\frac{-\pi r^2 R}{R_{ring}}\right) \cdot B \cdot 2\pi r \)

After simplifying, we can find that:

• \( \mu \geq \frac{Qr^2R^2}{2mg} \)

Final Result

Thus, the condition for the coefficient of friction \( \mu \) to ensure that the ring remains in equilibrium is:

• \( \mu > \frac{Qr^2R^2}{2mg} \)

This corresponds to option (1) from your list. By following these logical steps, we can see how the interplay of magnetic forces and friction leads to the necessary condition for equilibrium.