Magnetic field at the centre of a long current carrying solenoid is $8.28 \times 10^{- 4} T$ . If length of the solenoid is halved and number of turns is doubled, what will be the magnetic field near the ends of the solenoid?

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$4.14 \times 10^{- 4} T$

$8.28 \times 10^{- 4} T$

$16.56 \times 10^{- 4} T$

$24.12 \times 10^{- 4} T$

answer is C.

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Detailed Solution

At the centre $B = μ_{0} n i = μ_{0} \frac{N}{l} i$

Near the ends, $B^{'} = \frac{μ_{0}}{2} \frac{N^{'}}{l^{'}} i$

$\begin{array}{l} ∴ \frac{B^{'}}{B} = \frac{1}{2} (\frac{N^{'}}{N}) (\frac{l}{l^{'}}) = \frac{1}{2} (2) (2) = 2 \\ ∴ B^{'} = 2 \times 8.28 \times 10^{- 4} T = 16.56 \times 10^{- 4} T \end{array}$

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