JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 5)
A uniform magnetic field of $$2 \times 10^{-3} \mathrm{~T}$$ acts along positive $$Y$$-direction. A rectangular loop of sides $$20 \mathrm{~cm}$$ and $$10 \mathrm{~cm}$$ with current of $$5 \mathrm{~A}$$ is in $$Y-Z$$ plane. The current is in anticlockwise sense with reference to negative $$X$$ axis. Magnitude and direction of the torque is:
$$2 \times 10^{-4} \mathrm{~N}$$- $$\mathrm{m}$$ along negative $$Z$$-direction
$$2 \times 10^{-4} \mathrm{~N}$$ - $$\mathrm{m}$$ along positive $$X$$-direction
$$2 \times 10^{-4} \mathrm{~N}$$ - $$\mathrm{m}$$ along positive $$Y$$-direction
$$2 \times 10^{-4} \mathrm{~N}$$ - $$\mathrm{m}$$ along positive $$Z$$-direction
Explanation
_31st_January_Evening_Shift_en_5_1.png)
$$\begin{aligned} & \overrightarrow{\mathrm{M}}=\mathrm{i} \overrightarrow{\mathrm{A}} \\ & =5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}}) \\ & =0.1(-\hat{\mathrm{i}}) \\ & \vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}}) \\ & =2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m} \end{aligned}$$
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