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A charged particle projected in a magnetic field `B=(3hati+4hatj)xx10^-2T` The acceleration of the particle is found to be `a=(xhati+2hatj)m/s^2` find the value of `x` |
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Answer» As we have read e`F_m_|_B` i.e. the acceleration `a_|_B` or `a.b=0` or `(xhati+2hatj).(3hati+4hatj)xx10^=-2=0` `or (3x+8)xx10^-2=0` `x=-8/3m/s^2` |
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