A small charged ball having mass m and charge q is suspended from a rigid support by means of an inextensible thread of length l. It is made to rotate on a horizontal circular path in a uniform, time independent magnetic field of induction B which is directed upward. The time period of revolution of the ball is `T_0.` If the thread is always stretched, calculate the radius of circular path on which the ball moves.

Explanation

To solve the problem, we need to analyze the forces acting on the charged ball and derive the radius of the circular path it moves in. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a charged ball of mass \( m \) and charge \( q \) suspended by a thread of length \( l \). The ball rotates in a horizontal circular path in a uniform magnetic field \( B \) directed upward. The ball makes an angle \( \theta \) with the vertical.

### Step 2: Identify Forces Acting on the Ball
1. **Weight (mg)**: Acts downward.
2. **Tension (T)**: Acts along the thread, making an angle \( \theta \) with the vertical.
3. **Magnetic Force**: Acts perpendicular to the velocity of the ball, given by \( F_B = qvB \), where \( v \) is the linear velocity of the ball.

### Step 3: Resolve Forces
- Along the vertical direction (y-axis):
\[ T \cos \theta = mg \quad \text{(1)} \]
- Along the horizontal direction (x-axis):
\[ T \sin \theta = qvB \quad \text{(2)} \]

### Step 4: Relate Linear Velocity to Angular Velocity
The linear velocity \( v \) can be expressed in terms of the angular velocity \( \omega \) and the radius \( R \) of the circular path:
\[ v = \omega R \]

### Step 5: Substitute \( v \) in the Forces
Substituting \( v \) in equation (2):
\[ T \sin \theta = q(\omega R)B \]

### Step 6: Express Tension \( T \)
From equation (1):
\[ T = \frac{mg}{\cos \theta} \]
Substituting \( T \) in the horizontal force equation:
\[ \frac{mg}{\cos \theta} \sin \theta = q(\omega R)B \]

### Step 7: Simplify the Equation
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ mg \tan \theta = q(\omega R)B \]

### Step 8: Relate \( R \) and \( \theta \)
From the geometry of the problem, we have:
\[ R = l \sin \theta \]

### Step 9: Substitute \( R \) into the Equation
Substituting \( R \) into the equation:
\[ mg \tan \theta = q(\omega (l \sin \theta))B \]
This gives:
\[ mg \frac{R}{\sqrt{l^2 - R^2}} = q \omega (l \sin \theta) B \]

### Step 10: Solve for \( R \)
Squaring both sides and rearranging gives:
\[ R^2 = \frac{m^2 g^2}{(q \omega B)^2 + m^2 g^2} l^2 \]

### Final Result
Thus, the radius \( R \) of the circular path is given by:
\[ R = \sqrt{l^2 - \frac{m^2 g^2}{(q \omega B)^2 + m^2 g^2} l^2} \]