A proton is projected with a velocity 10^7, m | vedclass

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(B) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path.The time period $T$ of the circular motion is g

Vedclass · Accepted Answer

$1.62 	imes 10^{-7}$

Vedclass · Answer

(B) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path.The time period $T$ of the circular motion is given by $T = \frac{2\pi m}{qB}$.Here,$m = 1.65 	imes 10^{-27}\, kg$,$q = 1.6 	imes 10^{-19}\, C$,and $B = 100\, mT = 0.1\, T$.The time taken to traverse a $90^o$ arc is one-fourth of the total time period,i.e.,$t = \frac{T}{4} = \frac{2\pi m}{4qB} = \frac{\pi m}{2qB}$.Substituting the values: $t = \frac{3.14 	imes 1.65 	imes 10^{-27}}{2 	imes 1.6 	imes 10^{-19} 	imes 0.1}$.$t = \frac{5.181 	imes 10^{-27}}{0.32 	imes 10^{-19}} = 16.19 	imes 10^{-8} = 1.619 	imes 10^{-7}\, s$.Rounding to two decimal places,we get $t \approx 1.62 	imes 10^{-7}\, s$.

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

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