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(D) 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 t

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$\frac{2B}{3}$

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(D) 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 flowing through the solenoid.Given that the initial magnetic field is $B = \mu_0 n I$.According to the problem,the new number of turns per unit length $n' = 2n$ and the new current $I' = \frac{1}{3}I$.The new magnetic field $B'$ is given by:$B' = \mu_0 n' I'$$B' = \mu_0 (2n) \left(\frac{1}{3}I\right)$$B' = \frac{2}{3} (\mu_0 n I)$$B' = \frac{2}{3} B$Therefore,the new value of the magnetic field is $\frac{2B}{3}$.

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per $cm$ is doubled and the current is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the new value of the magnetic field will be:

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