2. Solved Papers
  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    A charged particle with velocity\[2\times {{10}^{3}}\]m/s passes undeflected through electric and magnetic field is 1.5 T. The electric field intensity would be

    A)  \[2\times {{10}^{3}}N/C\]     

    B)  \[1.5\times {{10}^{3}}N/C\]

    C)  \[3\times {{10}^{3}}N/C\]     

    D)  \[\frac{4}{3}\times {{10}^{-3}}N/C\]

    Correct Answer: C

    Solution :

     Given\[V=2\times {{10}^{3}}\]and \[B=1.5\text{ }T\] As,             \[qvB=qE\]        \[E=vB\] \[E=2\times {{10}^{3}}\times 1.5\] \[E=3\times {{10}^{3}}N/C\]


You need to login to perform this action.
You will be redirected in 3 sec spinner