A proton is moving with a uniform velocity of | vedclass

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

Question

(A) Given:Velocity of proton $v = 10^{6} \ m/s$ along the $Y$-axis.Electric field $E = 2 	imes 10^{4} \ V/m$ along the negative $X$-axis.Specific cha

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) Given:Velocity of proton $v = 10^{6} \ m/s$ along the $Y$-axis.Electric field $E = 2 	imes 10^{4} \ V/m$ along the negative $X$-axis.Specific charge $\frac{e}{m} = 10^{8} \ C/kg$.Since the proton moves with a uniform velocity, the net force on it is zero. Thus, the magnetic force balances the electric force:$qE = qvB$$B = \frac{E}{v} = \frac{2 	imes 10^{4}}{10^{6}} = 2 	imes 10^{-2} \ T$When the electric field is switched off, the proton moves in a circular path due to the magnetic force acting as the centripetal force:$R = \frac{mv}{qB} = \frac{v}{(e/m)B}$$R = \frac{10^{6}}{10^{8} 	imes 2 	imes 10^{-2}}$$R = \frac{10^{6}}{2 	imes 10^{6}} = 0.5 \ m$Therefore, the radius of the circle is $0.5 \ m$.

Similar Questions

$A$ particle of mass $M$ and charge $Q$ moving with velocity $\vec{v}$ describes a circular path of radius $R$ when subjected to a uniform transverse magnetic field of induction $B$. The work done by the field when the particle completes one full circle is

The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

$A$ charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module