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(B) The magnetic field on the axis of a long solenoid is given by $B = \mu_{0} nI$,where $n = \frac{N}{\ell}$ is the number of turns per unit length.W

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(B) The magnetic field on the axis of a long solenoid is given by $B = \mu_{0} nI$,where $n = \frac{N}{\ell}$ is the number of turns per unit length.When the solenoid is cut into four equal parts,the length of each piece becomes $\ell^{\prime} = \frac{\ell}{4}$.The problem states that half the original number of turns are wound on this piece,so $N^{\prime} = \frac{N}{2}$.The new number of turns per unit length is $n^{\prime} = \frac{N^{\prime}}{\ell^{\prime}} = \frac{N/2}{\ell/4} = \frac{4N}{2\ell} = 2 \left(\frac{N}{\ell}\right) = 2n$.Since the current $I$ remains the same,the new magnetic field $B^{\prime}$ is:$B^{\prime} = \mu_{0} n^{\prime} I = \mu_{0} (2n) I = 2(\mu_{0} nI) = 2B$.

$A$ long solenoid produces a magnetic field $B$ on its axis. If it is cut into four equal parts and,for the same current,half the number of turns are wound on any one piece,the value of the magnetic field on its axis becomes:

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