A particle of charge q and mass m enters normally (at point P) in a region of magnetic field with speed v. It comes out normally from Q after time T as shown in fig. The magnetic field B is present only in the region of radius R and is uniform. Initial and final velocities are along radial direction and they are perpendicular to each other. For this to happen, which of the following expression(s) is/are correct?

$B = \frac{m v}{q R}$
$T = \frac{π R}{2 v}$
none of these
$T = \frac{π m}{2 q B}$

Question

A particle of charge q and mass m enters normally (at point P) in a region of magnetic field with speed v. It comes out normally from Q after time T as shown in fig. The magnetic field B is present only in the region of radius R and is uniform. Initial and final velocities are along radial direction and they are perpendicular to each other. For this to happen, which of the following expression(s) is/are correct?

$B = \frac{m v}{q R}$
$T = \frac{π R}{2 v}$
none of these
$T = \frac{π m}{2 q B}$

Toppr Admin · Accepted Answer

The particle will move along an arc which is path of a circle of radius-xA0-r-mvBqFrom fig- we can see that r-R-x2234-R-mvBqT-x3C0-r-2v-x3C0-r2v-x3C0-R2v-x2235-r-R-mvBq-x2234-T-x3C0-m2Bq

The particle will move along an arc which is path of a circle of radius: r=mvBqFrom fig, we can see that r=R∴R=mvBqT=πr/2v=πr2v=πR2v(∵r=R=mvBq)∴T=πm2Bq

The particle will move along an arc which is path of a circle of radius:

$r = \frac{m v}{B q}$

From fig, we can see that $r = R$

$∴ R = \frac{m v}{B q}$

$T = \frac{π r / 2}{v} = \frac{π r}{2 v} = \frac{π R}{2 v} (∵ r = R = \frac{m v}{B q})$

$∴ T = \frac{π m}{2 B q}$