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WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer
    A straight conductor 0.1 m long moves in a  uniform magnetic field 0.1 T. The velocity of  the conductor is 15 m/s and is directed  perpendicular to the field. The emf induced  between the two ends of the conductor is

    A)  0.10V 

    B) 0.15V

    C)  1.50 V

    D)  15.00 V

    Correct Answer: B

    Solution :

      Given, length of conductor \[l=0.1\,m\] Mangetic field, \[B=0.1\,T\] Velocity of conductor, \[v=15\,m/s\] The angle between v and b is \[{{90}^{o}}\] When v and b are mutually perpendicular,, then emf (induced) is given by \[\varepsilon =vBl=15\times 0.1\times 0.1=\frac{15}{100}=0.15\,V\]


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