JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 12)
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is :
$$\mathrm{1~A~m^{-1}}$$
$$\mathrm{2.4\times10^{-3}~A~m^{-1}}$$
$$\mathrm{1.2\times10^{3}~A~m^{-1}}$$
$$\mathrm{2.4\times10^{3}~A~m^{-1}}$$
Explanation
Number of turns per unit length $=\frac{1200}{2}=600$
So, Magnetic Intensity $H=n I$
$$ \begin{aligned} & =600 \times 2 ~\mathrm{Am}^{-1} \\\\ & =1200 ~\mathrm{Am}^{-1} \end{aligned} $$
So, Magnetic Intensity $H=n I$
$$ \begin{aligned} & =600 \times 2 ~\mathrm{Am}^{-1} \\\\ & =1200 ~\mathrm{Am}^{-1} \end{aligned} $$
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