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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 14

  • question_answer
    A cyclotron is operated at an oscillator frequency of 24 MHz and has a dee radius R = 60 cm. What is magnitude of the magnetic field B (in tesla) to accelerate deuterons (mass\[=3.34\times {{10}^{-27}}kg\])?

    A) 9.5       

    B) 7.2   

    C) 5.0       

    D) 3.2

    Correct Answer: D

    Solution :

    [d]: Given : \[\upsilon =24\times {{10}^{6}}Hz,R=0.60\text{ }m\] We know that, \[R=\frac{mv}{qB}\] \[\therefore \]\[B=\frac{mv}{qR}\]                                                    ...(i) where \[v=\omega R=2\pi \upsilon R\] \[B=\frac{(3.34\times {{10}^{-27}})(9.04\times {{10}^{7}})}{1.6\times {{10}^{-19}}\times 0.60}=3.2T\]


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