Uniform magnetic fields of different strengths ( and ) both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m and charge q, at the interface at an instant, moves into the region 2 with velocity v and returns to th

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Uniform magnetic fields of different strengths ( $B_{1}$ and $B_{2}$ ) both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m and charge q, at the interface at an instant, moves into the region 2 with velocity v and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface

(Consider the velocity of the particle to be normal to the magnetic field and $B_{2} > B_{1}$ ) [2025]

1 $\frac{m v}{q B_{1}} (1 - \frac{B_{2}}{B_{1}}) \times 2$

2 $\frac{m v}{q B_{1}} (1 - \frac{B_{1}}{B_{2}})$

3 $\frac{m v}{q B_{1}} (1 - \frac{B_{2}}{B_{1}})$

4 $\frac{m v}{q B_{1}} (1 - \frac{B_{1}}{B_{2}}) \times 2$

Ans.
(4)

As $\vec{v}$ is $⊥$ to $\vec{B}$ , so charge particle will move in circular path, whose radius is given by

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

Starting point $\to A$

Ending point $\to C$

$∴$ Net displacement = AC

AC = CD – AD

$A C = \frac{2 m v}{q B_{1}} - \frac{2 m v}{q B_{2}}$

$A C = \frac{2 m v}{q B_{1}} [1 - \frac{B_{1}}{B_{2}}]$

Want Us to Email you About Special Offers & Updates?