JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    A charge q is spread uniformly over an insulated loop of radius. If it is rotated with an angular velocity\[\omega \] with respect to normal axis then the magnetic moment of the loop is [JEE Main 16-4-2018]

    A) \[A\frac{1}{2}q\omega {{r}^{2}}\]            

    B)  \[\frac{4}{3}q\omega {{r}^{2}}\]

    C)  \[\frac{3}{2}q\omega {{r}^{2}}\]             

    D) \[q\omega {{r}^{2}}\]

    Correct Answer: A

    Solution :

     Let us take an element at an angle \[\theta \] subtending an angle \[d\theta \] The charge the element has can be written as \[dq=\frac{q}{2\pi }d\theta \] We know that\[i=\frac{dq}{dt}=\frac{q}{2\pi }\times \frac{d\theta }{dt}\] The time dt can be written as \[dt=\frac{d\theta }{w}\] Hence\[i=\frac{qw}{2\pi }\] magnetic moment = iA Hence Magnetic moment\[M=\frac{qw}{2\pi }\times \pi {{r}^{2}}=\frac{qw{{r}^{2}}}{2}\]


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