JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    You are given a closed circuit with radii a and b as shown in fig carrying current i. The magnetic dipole moment of the circuit is

    A) \[\pi ({{a}^{2}}+{{b}^{2}})i\]                       

    B) \[\frac{1}{2}\pi ({{a}^{2}}+{{b}^{2}})i\]

    C) \[\pi ({{a}^{2}}-{{b}^{2}})i\]

    D) \[\frac{1}{2}\pi ({{a}^{2}}-{{b}^{2}})i\]

    Correct Answer: B

    Solution :

    m = current x area \[=i(\frac{1}{2}\pi {{a}^{2}}+\frac{1}{2}\pi {{b}^{2}})\] \[=\frac{1}{2}i\pi ({{a}^{2}}+{{b}^{2}})\]


You need to login to perform this action.
You will be redirected in 3 sec spinner