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Length of the solenoid is 60cm and its total number of turns are 1250. If 2A current is passed through it, what will be the magnetic field at any point on its axis 

Option: 1

5.23 \times 10^{-3} \mathrm{w b} / \mathrm{m}^2


Option: 2

6.23 \times 10^{-3} \mathrm{wb} / \mathrm{m}^2


Option: 3

2.03


Option: 4

None


Answers (1)

best_answer

Number of turns = 1250, length = 60cm = 0.6m

\therefore number of turns per unit length, n = \frac{1250}{0.6} m^{-1}

So, the magnetic field at any point on its axis,

\begin{aligned} B & =\mu_0 n I \\ & =4 \pi \times 10^{-7} \times \frac{1250}{0.6} \times 2 \\ & =5.23 \times 10^{-3} \mathrm{w b} / \mathrm{m}^2 \end{aligned}

Posted by

Ajit Kumar Dubey

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