JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2024
- Hard
- Numerical
Two charges of \(-4 \ \mu \mathrm{C}\) and \(+4\ \mu \mathrm{C}\) are placed at the points \(A(1,0,4) \mathrm{m}\) and \(B(2,-1,5) \mathrm{m}\) located in an electric field \(\vec{E}=0.20 \hat{\mathrm{i}} \mathrm{V} / \mathrm{cm}\). The magnitude of the torque acting on the dipole is \(8 \sqrt{\alpha} \times 10^{-3} \mathrm{Nm}\), Where \(\alpha=\)___________
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Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\( \vec{\tau}=\vec{p} \times \vec{E} \)
\( \vec{p}=q \vec{\ell} \)
\( \vec{E}=0.2 \frac{V}{\mathrm{~cm}}=20 \frac{\mathrm{V}}{\mathrm{m}} \)
\( \vec{p}=4 \times(\hat{i}-\hat{j}+\hat{k}) \)
\( =(4 \hat{i}-4 \hat{j}+4 \hat{k}) \mu \mathrm{C}-\mathrm{m} \)
\( \vec{\tau}=(4 \hat{i}-4 \hat{j}+4 \hat{k}) \times(20 \hat{i}) \times 10^{-6} \mathrm{Nm} \)
\( =(8 \hat{\mathrm{k}}+8 \hat{\mathrm{j}}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \)
\( \alpha=2\)
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