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A straight current (i) carrying conductor and having a mass m is hinged at a point on the boundary of magnetic field of intensity B as shown. The system lies in the x-y smooth horizontal plane. If the conductor is released, the angular acceleration of the conductor will be

Option: 1

\frac{3iB}{m}


Option: 2

\frac{3iB}{4m}

 


Option: 3

\frac{3iB}{2m}


Option: 4

\frac{3iB}{\sqrt{2}m}


Answers (1)

best_answer

dF=iBdl\\ \int d\tau = \int iBdl\:dl\\ \\ \tau =\frac{iBl^2}{2}\\ \\ \tau =l\alpha \\ \\ \alpha= \frac{iBl^3}{2ml^2}= \frac{3iB}{2m}

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Ritika Jonwal

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