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WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of \[{{60}^{o}}\] and the work done is W. The torque on the magnetic needle at this position is

    A)  \[2\sqrt{3}\,W\]

    B)  \[\sqrt{3}\,W\]

    C)  \[\frac{\sqrt{3}}{2}W\]

    D)  \[\frac{\sqrt{3}}{4}W\]

    Correct Answer: B

    Solution :

     Given, work done \[=W\]and \[\theta ={{60}^{o}}\] We know that \[W=MB(1-cos\theta )\] \[W=MB(1-cos{{60}^{o}})\] \[W=\frac{MB}{2}\] Hence,     \[|\tau |=MB\,sin{{60}^{o}}=\sqrt{3}\,W\] | t | = MB sin 60° = V3 W


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