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 (integer)
35
Solution
<p>$$\begin{aligned}
& B_H=3.5 \times 10^{-5} T \\
& F=i \ell B \sin \theta, \quad \mathrm{i}=\sqrt{2} \mathrm{~A} \\
& \frac{F}{\ell}=i B \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \\
& =35 \times 10^{-6} \mathrm{~N} / \mathrm{m}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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