Question:
The magnetic field intensity (H) at the centre of a long solenoid having 'n' turns per unit length and carrying a current I, when no material is kept in it is (\(μ_0\) = permeability of free space)
The magnetic field intensity (H) at the centre of a long solenoid having 'n' turns per unit length and carrying a current I, when no material is kept in it is (\(μ_0\) = permeability of free space)
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H equals turns per unit length times current.
Updated On: Oct 1, 2026
- \(n/I\)
- \(nI\)
- \(μ_0I\)
- \(μ_0/nI\)
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The Correct Option is B
Solution and Explanation
Step 1: Definition
For a long solenoid the magnetic field is \(B = \mu_0nI\). The magnetic intensity is \(H = \frac{B}{\mu_0} = nI\), for no material inside.
Step 2: Options
\(\frac nI\) and \(\frac{\mu_0}{nI}\) have wrong dimensions, and \(\mu_0I\) is missing \(n\). Option (B).
Final Answer:
H equals nI. \[ \boxed{\text{(B)}\ nI} \]
For a long solenoid the magnetic field is \(B = \mu_0nI\). The magnetic intensity is \(H = \frac{B}{\mu_0} = nI\), for no material inside.
Step 2: Options
\(\frac nI\) and \(\frac{\mu_0}{nI}\) have wrong dimensions, and \(\mu_0I\) is missing \(n\). Option (B).
Final Answer:
H equals nI. \[ \boxed{\text{(B)}\ nI} \]
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