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 creates an electric current. A charge revolving in a circle creates a continuous current loop, which subsequently produces a magnetic field at the center of the loop.
Step 2: Key Formula or Approach:
Equivalent current of a revolving charge: \(I = qf\), where \(f\) is frequency.
Magnetic field at the center of a circular current loop: \(B = \frac{\mu_0 I}{2r}\).
Step 3: Detailed Explanation:
Calculate the equivalent current:
Charge \(q = 10e = 10 \times 1.6 \times 10^{-19} \text{ C} = 1.6 \times 10^{-18} \text{ C}\).
Frequency \(f = 10 \text{ Hz}\) (rotations per second).
\[ I = qf = (1.6 \times 10^{-18}) \times 10 = 1.6 \times 10^{-17} \text{ A} \] Now, calculate the magnetic field \(B\):
Given radius \(r = 0.5 \text{ m}\).
\[ B = \frac{\mu_0 I}{2r} = \frac{(4\pi \times 10^{-7})(1.6 \times 10^{-17})}{2 \times 0.5} \] Since \(2 \times 0.5 = 1\):
\[ B = 4\pi \times 1.6 \times 10^{-24} \] Using \(\pi \approx 3.1415\):
\[ B \approx 4 \times 3.1415 \times 1.6 \times 10^{-24} \] \[ B \approx 12.56 \times 1.6 \times 10^{-24} \] \[ B \approx 20.1 \times 10^{-24} \text{ T} \] Adjust the scientific notation:
\[ B \approx 2.01 \times 10^{-23} \text{ T} \] This is approximately \(2 \times 10^{-23} \text{ T}\).
Step 4: Final Answer:
The magnetic field is approximately \(2 \times 10^{-23}T\).