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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    A radio transmitter transmits at 830 kHz. At a certain distance from the transmitter magnetic field has amplitude \[4.82\times {{10}^{-11}}T.\]The electric field and the wavelength are respectively   JEE Main Online Paper (Held On 26-May-2012)  

    A)  0.014N/C, 36m

    B)                        0.14N/C, 36m

    C)                         0.14N/C, 360m

    D)                         0.014N/C, 360m

    Correct Answer: D

    Solution :

                     Frequency of EM wave \[\upsilon =830\,KHz\]  \[=830\times {{10}^{3}}Hz.\] Magnetic field, \[B=4.82\times {{10}^{-11}}T\] As we know, frequency, \[\upsilon =\frac{c}{\lambda }\] or\[\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{830\times {{10}^{3}}}\]\[\lambda \simeq 360m\] And, \[E=BC=4.82\times {{10}^{-11}}\times 3\times {{10}^{8}}\] = 0.014 N/C


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