JEE Main & Advanced Sample Paper JEE Main - Mock Test - 1

  • question_answer
    In the figure shown ABC is a circle of radius a. Arc AB and AC each have resistance R. Arc BC has resistance 2R. A current I enters at point A and leaves the circle at B and C. All straight wires are radial. Calculate the magnetic field at the centre of the circle. Each arc AB, BC and AC subtends \[120{}^\circ \] at the centre of the circle.

    A) \[\frac{\sqrt{2}{{\mu }_{0}}I}{a}\]   

    B) \[\frac{{{\mu }_{0}}I}{8a}\]

    C) \[\frac{{{\mu }_{0}}I}{3a}\]            

    D) zero

    Correct Answer: D

    Solution :

    [d] Point B and C will have equal potential and there will be no current in arc BC. Current / gets divided into two equal parts at A. \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,{{B}_{AB}}=\frac{{{\mu }_{0}}(I/2)}{2a}\otimes \]             \[{{B}_{AC}}=\frac{{{\mu }_{0}}(I/2)}{2a}\odot \] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,{{B}_{centre}}=0\]


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