JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A uniform magnetic field of \(2 \times 10^{-3} \mathrm{~T}\) acts along positive \(\mathrm{Y}\)-direction. A rectangular loop of sides \(20\) \(\mathrm{cm}\) and \(10 \mathrm{~cm}\) with current of \(5 \mathrm{~A}\) is \(\mathrm{Y}-\mathrm{Z}\) plane. The current is in anticlockwise sense with reference to negative \(\mathrm{X}\) axis. Magnitude and direction of the torque is _______.
- A \(2 \times 10^{-4} \mathrm{~N}-\mathrm{m}\) along positive \(\mathrm{Z}\)-direction
- B \(2 \times 10^{-4} \mathrm{~N}-\mathrm{m}\) along negative \(Z\)-direction
- C \(2 \times 10^{-4} \mathrm{~N}-\mathrm{m}\) along positive \(\mathrm{X}\)-direction
- D \(2 \times 10^{-4} \mathrm{~N}-\mathrm{m}\) along positive \(\mathrm{Y}\)-direction
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-4} \mathrm{~N}-\mathrm{m}\) along negative \(Z\)-direction
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{M}}=\mathrm{iA}\) \(=5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}})\) \(=0.1(-\hat{\mathrm{i}})\) \(\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}})\)…
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