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NEET Sample Paper NEET Sample Test Paper-17

  • question_answer
    The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction \[1.5\times {{10}^{-2}}\]Tesla. A current of 10.0 ampere flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the lines of induction, then the torque acting on it is:

    A) \[~6000\text{ Newton }\!\!\times\!\!\text{ metre}\]

    B) \[\text{0 Newton}\,\text{ }\!\!\times\!\!\text{ }\,\text{metre}\]

    C) \[1.2\times {{10}^{-2}}\,\text{Newton}\,\text{ }\!\!\times\!\!\text{ }\,\text{metre}\]

    D) \[6\times {{10}^{-4}}\,\text{Newton }\!\!\times\!\!\text{ metre}\]

    Correct Answer: D

    Solution :

    torque \[\tau \]T acting on a current carrying coil of area A placed in a magnetic field of induction B is given by \[\tau =NIBA\sin \theta \] Here, \[N=1,B=1.5\times {{10}^{-2}}\]tesla \[A=0.05\times 0.08=40\times {{10}^{-4}}{{m}^{2}}\] \[I=10A,\theta ={{90}^{o}}=\pi /2.\] \[\therefore \]\[\tau =(1.5\times {{10}^{-2}})(10.0)(40\times {{10}^{-4}})sin\frac{\pi }{2}\] \[6\times {{10}^{-4}}\,Nm.\] Hence, the correction option is [d].


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