An electron is accelerated by a potential dif | vedclass

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

Question

(B) The kinetic energy $K$ gained by an electron accelerated through a potential difference $V$ is given by $K = qV$.The radius $r$ of the circular pa

Vedclass · Accepted Answer

$36.7\, cm$

Vedclass · Answer

(B) The kinetic energy $K$ gained by an electron accelerated through a potential difference $V$ is given by $K = qV$.The radius $r$ of the circular path of a charged particle in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB}$.Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2qV}{m}}$.Substituting $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Given values: $V = 12000\, V$,$B = 10^{-3}\, T$,$m = 9 	imes 10^{-31}\, kg$,$q = 1.6 	imes 10^{-19}\, C$.$r = \frac{1}{10^{-3}} \sqrt{\frac{2 	imes 9 	imes 10^{-31} 	imes 12000}{1.6 	imes 10^{-19}}} = 10^3 \sqrt{\frac{216 	imes 10^{-28}}{1.6 	imes 10^{-19}}} = 10^3 \sqrt{135 	imes 10^{-9}} = 10^3 \sqrt{1.35 	imes 10^{-7}} \approx 0.367\, m$.Therefore,$r = 36.7\, cm$.

Similar Questions

In a region of space,a uniform magnetic field $B$ exists in the $y-$direction. $A$ proton is fired from the origin,with its initial velocity $v$ making a small angle $\alpha$ with the $y-$direction in the $yz$ plane. In the subsequent motion of the proton,

An electron with energy $0.1 \, keV$ moves at a right angle to the Earth's magnetic field of $1 \times 10^{-4} \, Wb/m^2$. The frequency of revolution of the electron will be. (Take mass of electron $= 9.0 \times 10^{-31} \, kg$)

An electron and a proton of equal linear momentum enter in a direction perpendicular to a uniform magnetic field. If the radii of their circular paths are $r_e$ and $r_p$ respectively,then $\frac{r_e}{r_p}$ is equal to - (mass of electron $= m_e$,mass of proton $= m_p$)

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The angle of incidence is $45^{\circ}$. The radius of the circular portion of the path is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module