A charged ring of mass m = 50 gm, charge 2 coulomb and radius R = 2m is placed on a smooth horizontal surface. A magnetic field varying with time at a rate of (0.2 t) Tesla/sec is applied on to the ring in a direction normal to the surface of ring. Find the angular speed attained in a time `t_(1) = 10 sec`.

Answer

To solve the problem step by step, we will follow the principles of electromagnetic induction and rotational motion.

### Step 1: Understand the Given Information
- Mass of the ring, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Charge on the ring, \( Q = 2 \, \text{C} \)
- Radius of the ring, \( R = 2 \, \text{m} \)
- Rate of change of magnetic field, \( \frac{dB}{dt} = 0.2 \, \text{T/s} \)
- Time, \( t_1 = 10 \, \text{s} \)

### Step 2: Calculate the Area of the Ring
The area \( A \) of the ring is given by the formula:
\[ A = \pi R^2 \]
Substituting the value of \( R \):
\[ A = \pi (2)^2 = 4\pi \, \text{m}^2 \]

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A \]
Since \( B \) is changing with time, we can express it as:
\[ B(t) = B_0 + \frac{dB}{dt} \cdot t \]
Assuming \( B_0 = 0 \):
\[ B(t) = 0.2t \]
Thus, the magnetic flux becomes:
\[ \Phi(t) = (0.2t) \cdot (4\pi) = 0.8\pi t \, \text{Wb} \]

### Step 4: Calculate the Induced EMF
Using Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ \mathcal{E} = -\frac{d}{dt}(0.8\pi t) = -0.8\pi \, \text{V} \]

### Step 5: Calculate the Electric Field
The electric field \( E \) induced around the ring can be calculated using:
\[ E = \frac{\mathcal{E}}{2\pi R} \]
Substituting the values:
\[ E = \frac{-0.8\pi}{2\pi \cdot 2} = \frac{-0.8}{4} = -0.2 \, \text{V/m} \]

### Step 6: Calculate the Force on the Charge
The force \( F \) acting on the charge \( Q \) due to the electric field \( E \) is given by:
\[ F = Q \cdot E \]
Substituting the values:
\[ F = 2 \cdot (-0.2) = -0.4 \, \text{N} \]

### Step 7: Calculate the Torque
The torque \( \tau \) acting on the ring is given by:
\[ \tau = F \cdot R \]
Substituting the values:
\[ \tau = -0.4 \cdot 2 = -0.8 \, \text{N m} \]

### Step 8: Relate Torque to Angular Acceleration
Using the relation \( \tau = I \cdot \alpha \), where \( I \) is the moment of inertia of the ring:
\[ I = mR^2 = 0.05 \cdot (2)^2 = 0.2 \, \text{kg m}^2 \]
Thus, we have:
\[ -0.8 = 0.2 \cdot \alpha \implies \alpha = \frac{-0.8}{0.2} = -4 \, \text{rad/s}^2 \]

### Step 9: Calculate the Angular Speed
The angular speed \( \omega \) after time \( t_1 \) can be calculated using:
\[ \omega = \omega_0 + \alpha t_1 \]
Assuming the initial angular speed \( \omega_0 = 0 \):
\[ \omega = 0 + (-4)(10) = -40 \, \text{rad/s} \]
Since we are interested in the magnitude, we take:
\[ \omega = 40 \, \text{rad/s} \]

### Final Answer
The angular speed attained in a time \( t_1 = 10 \, \text{s} \) is:
\[ \omega = 40 \, \text{rad/s} \]