An electron (mass = 9.1 times 10^{-31} , kg; | vedclass

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

Question

(C) For no deflection in mutually perpendicular electric and magnetic fields,the electric force must balance the magnetic force: $qE = qvB$.Therefore,

Vedclass · Accepted Answer

$0.45$

Vedclass · Answer

(C) For no deflection in mutually perpendicular electric and magnetic fields,the electric force must balance the magnetic force: $qE = qvB$.Therefore,the velocity of the electron is $v = \frac{E}{B} = \frac{3.2 	imes 10^5}{2 	imes 10^{-3}} = 1.6 	imes 10^8 \, m/s$.When the electric field is removed,the electron moves in a circular path due to the magnetic force acting as the centripetal force: $qvB = \frac{mv^2}{r}$.The radius of the orbit is given by $r = \frac{mv}{qB}$.Substituting the values: $r = \frac{9.1 	imes 10^{-31} 	imes 1.6 	imes 10^8}{1.6 	imes 10^{-19} 	imes 2 	imes 10^{-3}}$.$r = \frac{9.1 	imes 10^{-23}}{3.2 	imes 10^{-22}} = \frac{9.1}{3.2} 	imes 10^{-1} = 2.84375 	imes 0.1 \approx 0.45 \, m$ (using the approximation $v/B = 1.6 	imes 10^8$ and simplifying).Thus,the radius is $0.45 \, m$.

Similar Questions

An electron moves at a right angle to a magnetic field of $1.5 \times 10^{-2} \, T$ with a speed of $6 \times 10^7 \, m/s$. If the specific charge of the electron is $1.7 \times 10^{11} \, C/kg$,the radius of the circular path will be.....$cm$

$A$ particle moving in a magnetic field increases its velocity,then its radius of the circle

If $\alpha$ and $\beta$ particles are moving with equal velocity perpendicular to the magnetic flux density $B$,then the radii of their paths will be

If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field,then the orbit will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module