JEE Main & Advanced AIEEE Solved Paper-2009

  • question_answer Directions Q. Nos. 13 are based on the following paragraph. A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius =b) and DA (radius = a) of the loop are joined by the straight wires AB and CD. A steady current \[I\] is flowing in the loop. Angle made by A B and CD at the origin O is 30°. Another straight thin wire with steady current \[{{I}_{2}}\] flowing out of the plane of the paper is kept at the origin. Due to the presence of the current \[{{I}_{1}}\] at the origin:     AIEEE  Solved  Paper-2009

    A) the forces on AB and DC are zero

    B) the forces on AD and BC are zero

    C) the magnitude of the net force on the loop is given by\[\frac{{{I}_{1}}I}{4\pi }{{\mu }_{0}}\left[ 2(b-a)+\frac{\pi }{3}(a+b) \right]\]

    D) the magnitude of the net force on the loop is given by\[\frac{{{\mu }_{0}}I{{I}_{1}}}{24ab}(b-a)\]

    Correct Answer: B

    Solution :

    \[\overrightarrow{F}=I(\overrightarrow{\ell }\times \overrightarrow{B})=0\]

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