JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 21)
The horizontal component of earth's magnetic field at a place is $$3.5 \times 10^{-5} \mathrm{~T}$$. A very long straight conductor carrying current of $$\sqrt{2} \mathrm{~A}$$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is __________ $$\times 10^{-6} \mathrm{~N} / \mathrm{m}$$.
Answer
35
Explanation
$$\begin{aligned}
& B_H=3.5 \times 10^{-5} T \\
& F=i \ell B \sin \theta, \quad \mathrm{i}=\sqrt{2} \mathrm{~A} \\
& \frac{F}{\ell}=i B \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \\
& =35 \times 10^{-6} \mathrm{~N} / \mathrm{m}
\end{aligned}$$
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