A proton is moving undeflected in a region of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $x = 2$.

Similar Questions

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

$A$ charged particle is moving in a uniform magnetic field. It penetrates a layer of lead and thereby loses half of its kinetic energy. What happens to the radius of curvature of its path?

An electron is moving along the $+x$ direction. To make it move along an anticlockwise circular path in the $x-y$ plane,the magnetic field must be applied along:

Vedclass Test Series

Exam Paper Generator

Online Exam Module