Question

A particle carrying a charge equal to 1000 times the charge on an electron, is rotating one rotation per second in a circular path of radius '$r$' m. If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum, the radius '$r$' in m is $[e = 1.6 \times 10^{-19} \text{C}] \quad [x = 2 \times 10^{-16}]$

• A. 0.04
• B. 0.02
• C. 0.2
• D. 0.4

Hint:

Current in a loop is charge times frequency of revolution ($I = q/T$).

Answer 1

The Correct Option is D

Step 1: Concept
Magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2r}$. Current $I = qf$, where $f$ is frequency.

Step 2: Meaning
$q = 1000e = 1000 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-16} \text{ C}$. $f = 1 \text{ rev/s}$, so $I = 1.6 \times 10^{-16} \text{ A}$.

Step 3: Analysis
Given $B = x \mu_0 = 2 \times 10^{-16} \mu_0$. Substituting in formula: $2 \times 10^{-16} \mu_0 = \frac{\mu_0 (1.6 \times 10^{-16})}{2r}$. $2 = 1.6 / 2r \implies 4r = 1.6$.

Step 4: Conclusion
$r = 0.4 \text{ m}$. Final Answer: (D)