Question

A particle of charge 10 times that of an electron is revolving in a circular path of radius $0.5 \text{ m}$. If the frequency of rotation is 10 rotations per second, the magnetic field at the centre of the circular path is approximately ($\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)}

• A. $4 \times 10^{-29} T$
• B. $2 \times 10^{-28} T$
• C. $2 \times 10^{-25} T$
• D. $8 \times 10^{-27} T$
• E. $2 \times 10^{-23} T

Hint:

Always convert frequency and charge into current ($I = qf$) before using magnetostatics formulas. It simplifies the problem into a standard current-loop case.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A moving charge behaves like a current loop. The magnetic field at the center of a circular current-carrying loop can be calculated using the Biot-Savart law for a circular path.
Key Formula or Approach:
1. Equivalent current: \( I = q \times f \).
2. Magnetic field at center: \( B = \frac{\mu_0 I}{2r} \).

Step 2: Detailed Explanation:
Given:
Charge \( q = 10 \times e = 10 \times 1.6 \times 10^{-19} \text{ C} = 1.6 \times 10^{-18} \text{ C} \)
Radius \( r = 0.5 \text{ m} \)
Frequency \( f = 10 \text{ Hz} \)
First, calculate the equivalent current:
\[ I = q \cdot f = (1.6 \times 10^{-18}) \times 10 = 1.6 \times 10^{-17} \text{ A} \]
Now, calculate the magnetic field:
\[ B = \frac{(4\pi \times 10^{-7}) \times (1.6 \times 10^{-17})}{2 \times 0.5} \]
The denominator \( 2 \times 0.5 = 1 \).
\[ B = 4\pi \times 1.6 \times 10^{-24} \]
\[ B \approx 12.56 \times 1.6 \times 10^{-24} \]
\[ B \approx 20.1 \times 10^{-24} = 2.01 \times 10^{-23} \text{ T} \]
Approximately, \( 2 \times 10^{-23} \text{ T} \).

Step 3: Final Answer:
The magnetic field is approximately $2 \times 10^{-23} T$.