A) \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega \cos \,(\omega t)\]
B) \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega \sin \,({{\omega }^{2}}{{t}^{2}})\]
C) \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega \sin \,(\omega t)\]
D) \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega {{\sin }^{2}}\,(\omega t)\]
Correct Answer: C
Solution :
We know that \[\tau =NBA\omega \sin \omega t\] Where \[N=\]number of loops \[=1\] \[B=\frac{{{\mu }_{0}}I}{2b}\]newton/amp-m \[A=\pi {{a}^{2}}metr{{e}^{2}}\] \[\therefore \] \[\tau =\frac{{{\mu }_{0}}I}{2b}(\pi {{a}^{2}})\omega sin\omega t\] \[=\frac{\pi {{a}^{2}}{{\mu }_{0}}I}{2b}.\omega \sin \omega t\]You need to login to perform this action.
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