A particle of mass m and charge q moves with | vedclass

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

Question

(B) The particle moves in a straight line until it reaches $x = a$. Upon entering the magnetic field region $(a \le x \le b)$,the magnetic force acts

Vedclass · Accepted Answer

$q(b - a)B/m$

Vedclass · Answer

(B) The particle moves in a straight line until it reaches $x = a$. Upon entering the magnetic field region $(a \le x \le b)$,the magnetic force acts as a centripetal force,causing the particle to move in a circular path of radius $r = \frac{mv}{qB}$.For the particle to just reach the region $x > b$,the radius of the circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.Therefore,the condition is $r \ge (b - a)$.Substituting the expression for $r$,we get $\frac{mv}{qB} \ge (b - a)$.Solving for $v$,we find $v \ge \frac{q(b - a)B}{m}$.Thus,the minimum velocity required is $v_{\min} = \frac{q(b - a)B}{m}$.

Similar Questions

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

The ratio of periods of $\alpha$-particle and proton moving on circular path in uniform magnetic field is . . . . . . .

Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to the magnetic field?

$A$ particle having charge $q$ enters a region of uniform magnetic field $\vec{B}$ (directed inwards) and is deflected a distance $x$ after travelling a distance $y$. The magnitude of the momentum of the particle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module