JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Two charges each of magnitude \(0.01\,C\) and separated by a distance of \(0.4\,mm\) constitute an electric dipole. If the dipole is placed in an uniform electric field ' \(\vec{E}\) ' of \(10\,dyne/C\) making \(30^{\circ}\) angle with \(\overrightarrow{ E }\), the magnitude of torque acting on dipole is :
- A \(4.0 \times 10^{-10}\,Nm\)
- B \(2.0 \times 10^{-10}\,Nm\)
- C \(1.0 \times 10^{-8}\,Nm\)
- D \(1.5 \times 10^{-9}\,Nm\)
Answer & Solution
Correct Answer
(B) \(2.0 \times 10^{-10}\,Nm\)
Step-by-step Solution
Detailed explanation
\(|\overrightarrow{ P }|= qd\) \(=0.01 \times 0.4 \times 10^{-3}\) \(=4 \times 10^{-6}\) \(|\vec{\tau}|=\operatorname{PE} \sin \theta\) \(=4 \times 10^{-6} \times 10 \times 10^{-5} \times \sin 30\) \(=4 \times 10^{-6-5+1} \times \frac{1}{2}\) \(=2 \times 10^{-10}\)
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