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WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of \[{{45}^{o}}\] with the magnetic field, the applied force on the-wire will be

    A)  1.5 N             

    B)  3N

    C)  \[3\sqrt{2}\,N\]

    D)  \[\frac{3}{\sqrt{2}}N\]

    Correct Answer: D

    Solution :

     Given \[i=10\,A,B=0.15\,T,\theta ={{45}^{o}}\]and \[l=2m\] Here, \[F=ilB\sin \theta \] \[=10\times 2\times 0.15\,\sin {{45}^{o}}\] \[=\frac{3}{\sqrt{2}}N\]


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