Question

A charged particle moving along a straight-line path enters a uniform magnetic field of \( 4 \) mT at right angles to the direction of the magnetic field. If the specific charge of the charged particle is \( 8 \times 10^7 \) C/kg, the angular velocity of the particle in the magnetic field is:

• A. \( 64 \times 10^4 \) rad \( s^{-1} \)
• B. \( 32 \times 10^4 \) rad \( s^{-1} \)
• C. \( 16 \times 10^4 \) rad \( s^{-1} \)
• D. \( 48 \times 10^4 \) rad \( s^{-1} \)

Hint:

For charged particles in a uniform magnetic field, angular velocity is given by \( \omega = \frac{qB}{m} \).

Answer 1

The Correct Option is B

Step 1: Apply Angular Velocity Formula The angular velocity is given by: \[ \omega = q B / m \] Step 2: Compute Angular Velocity \[ \omega = (8 \times 10^7) \times (4 \times 10^{-3}) \] \[ \omega = 32 \times 10^4 \text{ rad } s^{-1} \] Thus, the correct answer is \( 32 \times 10^4 \) rad \( s^{-1} \).

Answer 2

Given:
- Magnetic field strength, \( B = 4 \, \text{mT} = 4 \times 10^{-3} \, \text{T} \)
- Specific charge of the particle, \( \frac{q}{m} = 8 \times 10^{7} \, \text{C/kg} \)

We need to find the angular velocity \( \omega \) of the charged particle moving perpendicular to the magnetic field.

Step 1: Recall the formula for angular velocity of a charged particle in a magnetic field:
\[ \omega = \frac{q B}{m} \] where \( q \) is the charge and \( m \) is the mass of the particle.

Step 2: Substitute the specific charge and magnetic field:
\[ \omega = \left( \frac{q}{m} \right) B = 8 \times 10^{7} \times 4 \times 10^{-3} = 32 \times 10^{4} \, \text{rad/s} \]

Therefore, the angular velocity of the particle in the magnetic field is:
\[ \boxed{32 \times 10^{4} \, \text{rad/s}} \]