Question 132 of 276

A charge particle of $2\mu C$ accelerated by a potential difference of $100V$ enters a region of uniform magnetic field of magnitude $4mT$ at right angle to the direction of field. The charge particle completes semicircle of radius $3cm$ inside magnetic field. The mass of the charge particle is ______ $\times {10}^{-18}kg$.

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Answer: 144

As we know, radius of the circular path is given by,
$r=\dfrac{mv}{qB} =\dfrac{\sqrt{2mK}}{qB} =\dfrac{\sqrt{2\times m\times qV}}{qB}$
$\Rightarrow {r}^{2}=\dfrac{2mV}{q{B}^{2}}$
$\Rightarrow m=\dfrac{{r}^{2}q{B}^{2}}{2V} =\dfrac{{\left(3\times {10}^{-2}\right)}^{2}\times \left(2\times {10}^{-6}\right)\times {\left(4\times {10}^{-3}\right)}^{2}}{2\times 100} =144\times {10}^{-18}kg$