A particle of mass m and charge moves with a constant velocity along the positive -direction. It enters a region containing a uniform magnetic field directed along the negative -direction, extending from to . The minimum value of required so that the

Question

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]

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(2)

When a charged particle is moving in a uniform magnetic field normal to its motion,

$q v B = \frac{m v^{2}}{R}$

Width of the magnetic field region $(b - a) \leq R,$ where $' R'$ is its radius of curvature inside the magnetic field.

$∴$ $R = \frac{m v}{q B} \geq (b - a)$ $\Rightarrow v_{\min} = \frac{(b - a) q B}{m}$

Solution

1 $\frac{q b B}{m}$

2 $\frac{q (b - a) B}{m}$

3 $\frac{q a B}{m}$

4 $\frac{q (b + a) B}{2 m}$

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Solution

1 qbBm

2 q(b-a)Bm

3 qaBm

4 q(b+a)B2m

Ans. (2) When a charged particle is moving in a uniform magnetic field normal to its motion, qvB=mv2R Width of the magnetic field region (b-a)≤R, where 'R' is its radius of curvature inside the magnetic field. ∴ R=mvqB≥(b-a)⇒vmin=(b-a)qBm

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1 $\frac{q b B}{m}$

2 $\frac{q (b - a) B}{m}$

3 $\frac{q a B}{m}$

4 $\frac{q (b + a) B}{2 m}$