A mass spectrometer is a device which selects | vedclass

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

Question

(C) The speed of the charged particle just before entering the magnetic field is $v_0$.By the work-energy theorem,the kinetic energy gained is equal t

Vedclass · Accepted Answer

$\frac{qB^2d^2}{8V}$

Vedclass · Answer

(C) The speed of the charged particle just before entering the magnetic field is $v_0$.By the work-energy theorem,the kinetic energy gained is equal to the work done by the electric field:$qV = \frac{1}{2}mv_0^2$Solving for $v_0$:$v_0 = \sqrt{\frac{2qV}{m}}$The radius $r$ of the circular path in the magnetic field is given by:$r = \frac{mv_0}{qB}$From the figure,the distance $d$ between the holes is the diameter of the circular path,so $r = \frac{d}{2}$.Substituting this into the radius formula:$\frac{d}{2} = \frac{m}{qB} \sqrt{\frac{2qV}{m}}$Squaring both sides:$\frac{d^2}{4} = \frac{m^2}{q^2B^2} \cdot \frac{2qV}{m}$Simplifying the expression:$\frac{d^2}{4} = \frac{2mV}{qB^2}$Solving for $m$:$m = \frac{qB^2d^2}{8V}$

Similar Questions

The figure shows a circular coil carrying current $i$ kept very close but not touching at a point $A$ on a straight conductor carrying the same current $i$. The magnitude of magnetic induction at the centre $O$ of the circular coil will be:

Two infinite line-charges parallel to each other are moving with a constant velocity $v$ in the same direction as shown in the figure. The separation between two line-charges is $d$. The magnetic attraction balances the electric repulsion when,[$c$ = speed of light in free space]

The net magnetic field at the centre $O$ of the circle due to the current-carrying loop as shown in the figure is $(\theta < 180^\circ)$.

Mark the correct statement.

Current flows through uniform,square frames as shown. In which case is the magnetic field at the centre of the frame not zero?

Vedclass Test Series

Exam Paper Generator

Online Exam Module