A charged particle is moving in a circular or | vedclass

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

Question

(B) Given:Speed $v = 3 	imes 10^6 \, m/s$Magnetic field $B = 2 	imes 10^{-4} \, T$Radius $R = 6 \, cm = 6 	imes 10^{-2} \, m$For a charged particle

Vedclass · Accepted Answer

$2.5 	imes 10^{11} \, C/kg$

Vedclass · Answer

(B) Given:Speed $v = 3 	imes 10^6 \, m/s$Magnetic field $B = 2 	imes 10^{-4} \, T$Radius $R = 6 \, cm = 6 	imes 10^{-2} \, m$For a charged particle moving in a circular path in a magnetic field,the magnetic Lorentz force provides the necessary centripetal force:$Bqv = \frac{mv^2}{R}$Rearranging the formula to find the charge to mass ratio $(q/m)$:$\frac{q}{m} = \frac{v}{BR}$Substituting the given values:$\frac{q}{m} = \frac{3 	imes 10^6}{(2 	imes 10^{-4}) 	imes (6 	imes 10^{-2})}$$\frac{q}{m} = \frac{3 	imes 10^6}{12 	imes 10^{-6}}$$\frac{q}{m} = 0.25 	imes 10^{12} \, C/kg = 2.5 	imes 10^{11} \, C/kg$Thus,the correct option is $B$.

$A$ charged particle is moving in a circular orbit of radius $6 \, cm$ with a uniform speed of $3 \times 10^6 \, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4} \, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is

Similar Questions

If a proton enters perpendicularly into a magnetic field with velocity $v$,the time period of revolution is $T$. If the proton enters with velocity $2v$,what will be the time period?

An electron (charge $q$ $C$) enters a magnetic field of $B$ $Wb/m^2$ with a velocity of $v$ $m/s$ in the same direction as that of the field. The force on the electron is:

$A$ neutron,a proton,an electron,and an $\alpha$-particle enter a region of uniform magnetic field with the same velocities. The magnetic field is perpendicular and directed into the plane of the paper. The tracks of the particles are labeled in the figure. Which track does the electron follow?

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be:

Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

Vedclass Test Series

Exam Paper Generator

Online Exam Module