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 $c m$ is halved, the new value of the magnetic field is -

B / 2

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B

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2 B

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4 B

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Solution

The correct option is C $B$
$B = μ_{0} n i; n^{'} = \frac{n}{2}; i^{'} = 2 i \Rightarrow B^{'} = B$

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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 -

The correct option is C BB=μ0ni;n′=n2;i′=2i⇒B′=B

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 $c m$ is halved, the new value of the magnetic field is -

The correct option is C $B$
$B = μ_{0} n i; n^{'} = \frac{n}{2}; i^{'} = 2 i \Rightarrow B^{'} = B$