JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 8)
A rigid wire consists of a semicircular portion of radius $$R$$ and two straight sections. The wire is partially immerged in a perpendicular magnetic field $$B=B_0 \hat{k}$$ as shown in figure. The magnetic force on the wire if it has a current $$i$$ is:
_31st_January_Morning_Shift_en_8_1.png)
$$i B R \hat{j}$$
$$-2 i B R \hat{j}$$
$$2 i B R \hat{j}$$
$$-i B R \hat{j}$$
Explanation
_31st_January_Morning_Shift_en_8_2.png)
Note : Direction of magnetic field is in $$+\hat{k}$$
$$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{i} \vec{\ell} \times \overrightarrow{\mathrm{B}} \\ & \ell=2 \mathrm{R} \\ & \overrightarrow{\mathrm{F}}=-2 \mathrm{iRB} \hat{\mathrm{j}} \end{aligned}$$
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