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(A) When the proton moves undeflected,the electric force equals the magnetic force: $eE = evB$,which gives $E = vB$ or $B = E/v$. When the electric fi

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) When the proton moves undeflected,the electric force equals the magnetic force: $eE = evB$,which gives $E = vB$ or $B = E/v$. When the electric field is switched off,the proton moves in a circular path of radius $R$ due to the magnetic field: $R = \frac{mv}{eB}$. Substituting $B = E/v$ into the radius formula: $R = \frac{mv}{e(E/v)} = \frac{mv^2}{eE}$. Rearranging to solve for $E$: $E = \frac{mv^2}{eR}$. Given: $m = 1.6 	imes 10^{-27} 	ext{ kg}$,$v = 2 	imes 10^5 	ext{ ms}^{-1}$,$R = 2 	ext{ cm} = 0.02 	ext{ m}$,$e = 1.6 	imes 10^{-19} 	ext{ C}$. $E = \frac{(1.6 	imes 10^{-27}) 	imes (2 	imes 10^5)^2}{(1.6 	imes 10^{-19}) 	imes 0.02} = \frac{1.6 	imes 10^{-27} 	imes 4 	imes 10^{10}}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-2}} = \frac{4 	imes 10^{-17}}{2 	imes 10^{-21}} = 2 	imes 10^4 	ext{ N/C}$. Comparing this with $x 	imes 10^4 	ext{ N/C}$,we get $x = 2$.

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