An electron of mass m and charge e is project | vedclass

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

Question

(A) The electron moves in a circular path of radius $R$ in the magnetic field $B$. The radius of the path is given by $R = \frac{mv_0}{eB}$.From the g

Vedclass · Accepted Answer

$B = \frac{2bmv_0}{(b^2 - a^2)e}$

Vedclass · Answer

(A) The electron moves in a circular path of radius $R$ in the magnetic field $B$. The radius of the path is given by $R = \frac{mv_0}{eB}$.From the geometry of the problem,consider the right-angled triangle formed by the centre of the sphere,the point of projection,and the centre of the circular path of the electron.The distance from the centre of the sphere to the centre of the circular path is $(b - R)$. The radius of the sphere is $a$,and the radius of the circular path is $R$. Since the electron is projected normally,the radius of the sphere at the point of projection is perpendicular to the velocity $v_0$,which is tangent to the circular path.Applying the Pythagorean theorem to the triangle with hypotenuse $(b - R)$,base $a$,and height $R$ is not correct here. Instead,the triangle has sides $a$,$R$,and hypotenuse $(b - R)$ is incorrect. Looking at the geometry,the distance from the centre of the sphere to the wall is $b$. The centre of the circular path lies on the $x$-axis. The distance from the centre of the sphere to the centre of the circular path is $(b - R)$.In the right triangle formed by the centre of the sphere,the point of projection,and the centre of the circular path,the hypotenuse is $(b - R)$,one leg is $a$,and the other leg is $R$. Thus,$(b - R)^2 = a^2 + R^2$.$b^2 + R^2 - 2bR = a^2 + R^2$$b^2 - a^2 = 2bR$$R = \frac{b^2 - a^2}{2b}$Substituting $R = \frac{mv_0}{eB}$,we get $\frac{mv_0}{eB} = \frac{b^2 - a^2}{2b}$.Solving for $B$,we get $B = \frac{2bmv_0}{(b^2 - a^2)e}$.

Similar Questions

$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

$A$ proton, a deuteron, and an $\alpha$-particle are projected perpendicular to the direction of a uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is

An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module