Question

A particle of charge equal to 10 times the charge of an electron revolves in a circle with frequency equal to 10 revolutions per second. Find the magnetic field at the centre of the circular path.

Hint:

To find the magnetic field at the center of a current-carrying circular loop, use the formula \( B = \frac{\mu_0 I}{2r} \), where the current is calculated using the charge and frequency.

Answer 1

Step 1: Use the formula for the magnetic field due to a moving charge.
The magnetic field at the centre of a circular loop carrying current is given by the formula: \[ B = \frac{\mu_0 I}{2r} \] where:
- \( I \) is the current, and
- \( r \) is the radius of the loop.

Step 2: Calculate the current.
The current \( I \) is related to the charge and frequency by: \[ I = n q \] where:
- \( n = 10 \, \text{revolutions per second} \) is the frequency,
- \( q = 10 \times e \) is the charge of the particle, with \( e = 1.6 \times 10^{-19} \, \text{C} \).
Thus, the charge is: \[ q = 10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18} \, \text{C} \] Therefore, the current is: \[ I = 10 \times 1.6 \times 10^{-18} = 1.6 \times 10^{-17} \, \text{A} \]
Step 3: Use the radius of the circular path.
To find the magnetic field, we also need the radius of the circular path. For simplicity, we assume the radius is related to the particle’s velocity and the magnetic force. For this calculation, we can assume that the magnetic field is directly proportional to the current and charge as: \[ B = \frac{\mu_0 \times 1.6 \times 10^{-17}}{2r} \] Thus, the magnetic field at the center of the circular path is directly related to the current and radius of the circular path.