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(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_{0} n I$,where $n$ is the number of turns per unit length and $I$ i

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$B$

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(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_{0} n I$,where $n$ is the number of turns per unit length and $I$ is the current.Let the initial magnetic field be $B_{1} = \mu_{0} n I$.According to the problem,the new current $I' = 2I$ and the new number of turns per unit length $n' = \frac{n}{2}$.The new magnetic field $B_{2}$ is given by $B_{2} = \mu_{0} n' I' = \mu_{0} \left( \frac{n}{2} \right) (2I) = \mu_{0} n I$.Therefore,$B_{2} = B_{1} = B$.

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per $cm$ is halved,the new value of magnetic field will be equal to

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