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(D) For a particle moving in a circular path in a magnetic field,the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r} \Rightarrow

Vedclass · Accepted Answer

$2.0 	imes 10^{-24} \, kg$

Vedclass · Answer

(D) For a particle moving in a circular path in a magnetic field,the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r} \Rightarrow v = \frac{qBr}{m}$.When the particle moves in a straight path under the influence of both electric and magnetic fields,the net force is zero,meaning the electric force balances the magnetic force: $qE = qvB \Rightarrow E = vB$.Substituting the expression for $v$: $E = \left(\frac{qBr}{m}ight)B = \frac{qB^2r}{m}$.Rearranging for mass $m$: $m = \frac{qB^2r}{E}$.Given $q = 1.6 	imes 10^{-19} \, C$,$B = 0.5 \, T$,$r = 0.5 	imes 10^{-2} \, m$,and $E = 100 \, V/m$.$m = \frac{(1.6 	imes 10^{-19}) 	imes (0.5)^2 	imes (0.5 	imes 10^{-2})}{100} = \frac{1.6 	imes 10^{-19} 	imes 0.25 	imes 0.005}{100} = \frac{0.002 	imes 10^{-19}}{100} = 2.0 	imes 10^{-24} \, kg$.

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