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(D) The radius $r$ of a charged particle moving in a circular orbit in a magnetic field is given by the formula $r = \frac{mv}{qB}$.Since kinetic ener

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$25$

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(D) The radius $r$ of a charged particle moving in a circular orbit in a magnetic field is given by the formula $r = \frac{mv}{qB}$.Since kinetic energy $E = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mE}$.Substituting this into the radius formula: $r = \frac{\sqrt{2mE}}{qB}$.Given values: $m = 9 	imes 10^{-31} \, kg$,$E = 7.2 	imes 10^{-18} \, J$,$q = 1.6 	imes 10^{-19} \, C$,and $B = 9 	imes 10^{-5} \, Wb/m^2$.$r = \frac{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 7.2 	imes 10^{-18}}}{1.6 	imes 10^{-19} 	imes 9 	imes 10^{-5}}$$r = \frac{\sqrt{129.6 	imes 10^{-49}}}{14.4 	imes 10^{-24}} = \frac{\sqrt{12.96 	imes 10^{-48}}}{14.4 	imes 10^{-24}}$$r = \frac{3.6 	imes 10^{-24}}{14.4 	imes 10^{-24}} = \frac{3.6}{14.4} = 0.25 \, m$.Converting to centimeters: $0.25 \, m = 25 \, cm$.

An electron (mass $= 9 \times 10^{-31} \, kg$,charge $= 1.6 \times 10^{-19} \, C$) with a kinetic energy of $7.2 \times 10^{-18} \, J$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \, Wb/m^2$. The radius of the orbit is ..... $cm$.

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