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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer
    A beam of proton with velocity\[4\times {{10}^{5}}m/s\]enters a uniform magnetic field of 0.3 T at an angle of \[60{}^\circ \]to the magnetic field. Find the radius of the helical path taken by the proton beam.

    A)  0.2 cm          

    B)  1.2 cm

    C)  2.2 cm          

    D)  0.122 cm

    Correct Answer: B

    Solution :

     \[r=\frac{mv\sin \theta }{qB}\] \[r=\frac{1.67\times {{10}^{-27}}\times 4\times {{10}^{5}}\times \left( \frac{\sqrt{3}}{2} \right)}{1.6\times {{10}^{-19}}\times 0.3}\] \[=0.012m\] \[r=1.2cm\]


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