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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m \[{{s}^{-1}}\], at right angles to the horizontal component of the earths magnetic field of \[0.3\times {{10}^{-4}}Wb/{{m}^{2}}.\] The value of the induced emf in wire is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[0.3\times {{10}^{-3}}V\]                               

    B) \[2.5\times {{10}^{-3}}V\]

    C) \[1.5\times {{10}^{-3}}V\]                   

    D)   \[1.1\times {{10}^{-3}}V\]

    Correct Answer: C

    Solution :

    The motional emf is given as \[\varepsilon =\left| v(\overrightarrow{l}\times \overrightarrow{B}) \right|=\]\[5\times (0.3\times {{10}^{-4}})(10)sin{{90}^{o}}=1.5\times {{10}^{-3}}V\]


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