JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 23)
Two charges of $$-4 \mu \mathrm{C}$$ and $$+4 \mu \mathrm{C}$$ are placed at the points $$\mathrm{A}(1,0,4) \mathrm{m}$$ and $$\mathrm{B}(2,-1,5) \mathrm{m}$$ located in an electric field $$\overrightarrow{\mathrm{E}}=0.20 \hat{i} \mathrm{~V} / \mathrm{cm}$$. The magnitude of the torque acting on the dipole is $$8 \sqrt{\alpha} \times 10^{-5} \mathrm{Nm}$$, where $$\alpha=$$ _________.
Answer
2
Explanation
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$$\begin{aligned} & \vec{\tau}=\vec{p} \times \vec{E} \\ & \vec{p}=q \vec{\ell} \\ & \overrightarrow{\mathrm{E}}=0.2 \frac{\mathrm{V}}{\mathrm{cm}}=20 \frac{\mathrm{V}}{\mathrm{m}} \\ & \overrightarrow{\mathrm{p}}=4 \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mu \mathrm{C}-\mathrm{m} \\ & \vec{\tau}=(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(20 \hat{\mathrm{i}}) \times 10^{-6} \mathrm{Nm} \\ & =(8 \hat{\mathrm{k}}+8 \hat{\mathrm{j}}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \\ & \alpha=2 \end{aligned}$$
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