Question

A particle having a charge $100e$ is revolving in a circular path of radius $0.8\text{ m}$ with $1\text{ r.p.s.}$. The magnetic field produced at the centre of the circle in SI unit is ($\mu_0 = \text{permeability of vacuum}$, $e = 1.6 \times 10^{-19}\text{ C}$)

• A. $10^{-17}\mu_0$
• B. $10^{-3}\mu_0$
• C. $10^{-7}\mu_0$
• D. $10^{-11}\mu_0$

Hint:

Notice that the denominator $2R = 2(0.8) = 1.6$ matches the numeric leading coefficient of the charge $1.6 \times 10^{-19}$ perfectly. They cancel out cleanly, leaving only the powers of ten: $100 \times 10^{-19} = 10^{-17}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A point charge of magnitude $q = 100e$ completes $f = 1$ revolution per second (r.p.s.) along a circle of radius $R = 0.8\text{ m}$. We need to evaluate the magnetic field induction $B$ generated at the center of this orbit.

Step 2: Key Formula or Approach:
A revolving charge sets up an effective electric current loop given by $I = q \cdot f$. The magnetic field at the center of a circular current loop of radius $R$ is given by: $$B = \frac{\mu_0 I}{2R}$$

Step 3: Detailed Explanation:
First, let's find the effective loop current $I$: $$I = (100e) \times 1 = 100 \times (1.6 \times 10^{-19}\text{ C}) \times 1\text{ s}^{-1} = 1.6 \times 10^{-17}\text{ A}$$ Now, substitute this effective current and the radius $R = 0.8\text{ m}$ into our field formula: $$B = \frac{\mu_0 \cdot (1.6 \times 10^{-17})}{2 \times 0.8}$$ $$B = \frac{\mu_0 \cdot (1.6 \times 10^{-17})}{1.6}$$ The factor $1.6$ cancels out perfectly from the numerator and denominator: $$B = 10^{-17}\mu_0$$

Step 4: Final Answer:
The magnetic field produced at the center is $10^{-17}\mu_0$, which corresponds to option (A).