A long solenoid carrying current produces a magnetic field B along its axis. If the number of turns in the solenoid is halved and current in it is doubled the new magnetic field will be :
The correct answer is (B) B.
The magnetic field inside a solenoid is given by
B = μ0nI, where
n is the number of turns per unit length. Halving turns and doubling current keeps
nI constant, so the field remains B.
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The correct answer is (B) B.
The magnetic field inside a long solenoid is given by:
B = μonI
Where n = number of turns per unit length and
I = current.
If the number of turns is halved (n/2) and current is doubled (2I), the new magnetic field becomes:
B’ = μ0
n/2 2I = μ0
nI = B
Answer: (B) B.
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