JEE MAIN - Physics (2024 - 6th April Morning Shift - No. 16)
An element $$\Delta l=\Delta x\hat{i}$$ is placed at the origin and carries a large current $$I=10 \mathrm{~A}$$. The magnetic field on the $$y$$-axis at a distance of $$0.5 \mathrm{~m}$$ from the elements $$\Delta x$$ of $$1 \mathrm{~cm}$$ length is:
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$$10 \times 10^{-8} \mathrm{~T}$$
$$8 \times 10^{-8} \mathrm{~T}$$
$$4 \times 10^{-8} \mathrm{~T}$$
$$12 \times 10^{-8} \mathrm{~T}$$
Explanation
$$\begin{aligned}
& B=\frac{u_0}{4 \pi} \frac{i d l \sin \theta}{r^2} \\
& \Rightarrow B=\frac{10^{-7} \times 10 \times 10 \quad \times 1}{\frac{1}{4}}=4 \times 10^{-8} \mathrm{~T}
\end{aligned}$$
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