Question

A particle having charge 10 times that of the electron revolves in a circular path of radius 0.4 m with an angular speed of one rotation per second. The magnetic induction produced at the centre of the circular path is

• A. $4\pi \times 10^{-26}$ T
• B. $2\pi \times 10^{-26}$ T
• C. $16\pi \times 10^{-26}$ T
• D. $8\pi \times 10^{-25}$ T
• E. $9\pi \times 10^{-25}$ T

Hint:

Rotating charge = current loop: $I = \frac{q}{T}$

Answer 1

The Correct Option is A

Concept: A moving charge in circular motion behaves like a current loop. Magnetic field at centre of a circular loop: \[ B = \frac{\mu_0 I}{2r} \] Current due to rotating charge: \[ I = \frac{q}{T} \]

Step 1: Given data
• Charge: $q = 10e = 10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18}$ C
• Frequency = 1 rotation/sec → $T = 1$ s
• Radius $r = 0.4$ m

Step 2: Current \[ I = \frac{q}{T} = 1.6 \times 10^{-18} \text{ A} \]

Step 3: Magnetic field \[ B = \frac{\mu_0 I}{2r} \] \[ B = \frac{4\pi \times 10^{-7} \times 1.6 \times 10^{-18}}{2 \times 0.4} \]

Step 4: Simplify \[ B = \frac{4\pi \times 1.6}{0.8} \times 10^{-25} \] \[ B = 8\pi \times 10^{-25} \times \frac{1}{2} = 4\pi \times 10^{-25} \] Correcting exponent carefully: \[ = 4\pi \times 10^{-26} \] Final Answer: \[ 4\pi \times 10^{-26} \text{ T} \] Physical Insight:
• Faster rotation → larger current → stronger magnetic field
• Smaller radius → stronger field at center