Question:
A long solenoid carrying a current produces a magnetic field '\(B\)' along its axis. If the currunt is tripled and the number of turns per cm is halved then new value of magnetic field will be
A long solenoid carrying a current produces a magnetic field '\(B\)' along its axis. If the currunt is tripled and the number of turns per cm is halved then new value of magnetic field will be
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B = mu0 n I inside a long solenoid.
Updated On: Oct 1, 2026
- \(\frac{B}{2}\)
- \(B\)
- \(\frac{3}{2}B\)
- \(3B\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The magnetic field along the axis inside a long solenoid is \(B = \mu_0nI\), where \(n\) is the number of turns per unit length.
Step 2: Apply the changes:
The current becomes \(3I\) and the turns per unit length become \(\frac n2\):
\[ B' = \mu_0\cdot\frac n2\cdot3I = \frac32\mu_0nI = \frac32B \]
Step 3: Why the other options are wrong.
\(\frac B2\) accounts only for the halved turns. \(3B\) accounts only for the tripled current. \(B\) would need the two changes to cancel, which is not the case since \(3\times\frac12 = \frac32\).
Final Answer:
The new field is \(\frac32B\), option (C). \[ \boxed{\frac{3}{2}B} \]
The magnetic field along the axis inside a long solenoid is \(B = \mu_0nI\), where \(n\) is the number of turns per unit length.
Step 2: Apply the changes:
The current becomes \(3I\) and the turns per unit length become \(\frac n2\):
\[ B' = \mu_0\cdot\frac n2\cdot3I = \frac32\mu_0nI = \frac32B \]
Step 3: Why the other options are wrong.
\(\frac B2\) accounts only for the halved turns. \(3B\) accounts only for the tripled current. \(B\) would need the two changes to cancel, which is not the case since \(3\times\frac12 = \frac32\).
Final Answer:
The new field is \(\frac32B\), option (C). \[ \boxed{\frac{3}{2}B} \]
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