Question

If the time period of an alpha particle rotating in a circular path of radius 2 fermi is 3.14 $\mu$s, then the magnetic field induced at the centre of the circle is nearly

• A. 48 $\mu$T
• B. 16 $\mu$T
• C. 32 $\mu$T
• D. 64 $\mu$T

Hint:

- For circular motion of charged particle, $B = \frac{\mu_0 I}{2R}$.
- Current $I = q/T$.
- Units: $\mu$s → s, fermi → m.
- Charge of alpha particle = 2e.
- Always convert units before substituting into formulas.

Answer 1

The Correct Option is A

1. Magnetic field at the center of circular current loop: $B = \frac{\mu_0 I}{2R}$.
2. Current $I = \frac{q}{T}$ where $q$ = charge of alpha particle, $T$ = time period.
3. Given: $T = 3.14~\mu s = 3.14 \times 10^{-6}$ s, $R = 2$ fermi = $2 \times 10^{-15}$ m, $q = 2e = 3.2 \times 10^{-19}$ C.
\[ I = \frac{3.2 \times 10^{-19}}{3.14 \times 10^{-6}} \approx 1.02 \times 10^{-13}~\text{A} \]
4. $B = \frac{4\pi \times 10^{-7} \cdot 1.02 \times 10^{-13}}{2 \cdot 2 \times 10^{-15}} \approx 48~\mu T$