Magnetics 1 Question 11
11. A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$-direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$-direction, extending from $x=a$ to $x=b$. The minimum value of $v$ required so that the particle can just enter the region $x>b$ is
$(2002,2 M)$
(a) $\frac{q b B}{m}$
(b) $\frac{q(b-a) B}{m}$
(c) $\frac{q a B}{m}$
(d) $\frac{q(b+a) B}{2 m}$
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Answer:
Correct Answer: 11. (b)
Solution:
- If $(b-a) \geq r$
( $r=$ radius of circular path of particle)
The particle cannot enter the region $x>b$.
So, to enter in the region $x>b$
$$ \begin{aligned} & r>(b-a) \text { or } \frac{m v}{B q}>(b-a) \\ & \text { or } v>\frac{q(b-a) B}{m} \end{aligned} $$
