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A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is
- 4B
- B/2
- B
- 2B
Correct answer: B
Solution
The magnetic field inside a solenoid is given by B = μ₀nI, where n is the number of turns per unit length and I is the current. If the current is doubled and the number of turns per cm is halved, the product nI remains the same, so the magnetic field B remains unchanged.
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