Question:
A particle carrying a charge equal to \(100\) times the charge of an electron is rotating one rotation per second in a circular path of radius \(0.8\) m. The value of magnetic field produced at the centre will be (\(\mu _{0}\) = permeability of free space)
A particle carrying a charge equal to \(100\) times the charge of an electron is rotating one rotation per second in a circular path of radius \(0.8\) m. The value of magnetic field produced at the centre will be (\(\mu _{0}\) = permeability of free space)
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A rotating charge acts as a circular current I = q f, and the field at the centre of a loop is mu_0I/2r.
Updated On: Nov 12, 2026
- \(10^{-7}/\mu _{0}\)
- \(\mu _{0}10^{-17}\)
- \(\mu _{0}10^{-6}\)
- \(\mu _{0}10^{-7}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
A charge going around a circle once per second is equivalent to a current \(I=qf\).
Step 2: Find the Current
\(q=100\times1.6\times10^{-19}=1.6\times10^{-17}\) C, \(f=1\) Hz, so \(I=1.6\times10^{-17}\) A.
Step 3: Calculation
\(B=\dfrac{\mu_{0}I}{2r}=\dfrac{\mu_{0}\times1.6\times10^{-17}}{2\times0.8}=\mu_{0}\times10^{-17}\) T.
Final Answer:
The magnetic field at the centre is mu_0 x 10-17. \[ \boxed{\text{(B) }\mu_{0}10^{-17}} \]
A charge going around a circle once per second is equivalent to a current \(I=qf\).
Step 2: Find the Current
\(q=100\times1.6\times10^{-19}=1.6\times10^{-17}\) C, \(f=1\) Hz, so \(I=1.6\times10^{-17}\) A.
Step 3: Calculation
\(B=\dfrac{\mu_{0}I}{2r}=\dfrac{\mu_{0}\times1.6\times10^{-17}}{2\times0.8}=\mu_{0}\times10^{-17}\) T.
Final Answer:
The magnetic field at the centre is mu_0 x 10-17. \[ \boxed{\text{(B) }\mu_{0}10^{-17}} \]
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