Solution:

Ans. Condition under which, the particle will continue moving along $x$-axis,

$$ \begin{align*} q E & =q v B \ v & =\frac{E}{B} \end{align*} $$

Trajectory becomes helical about the direction of magnetic field.

[CBSE Marking Scheme 2017] AI Q. 3. A charge $q$ of mass $m$ is moving with a velocity of $v$, at right angles to a uniform magnetic field $B$. Deduce the expression for the radius of the circular path it describes. $\mathrm{R}$ [Delhi Compt. I, III 2017]

Ans. Derivation of the expression for radius 2 Force experienced by charged particle in magnetic field $\vec{B}$,

$$ \vec{F}=q(\vec{v} \times \vec{B}) $$

As $v$ and $B$ are perpendicular, $F=q v B \quad 1 / 2$ This force is perpendicular to the direction of velocity and hence acts as centripetal force.

$$ \begin{align*} \frac{m v^{2}}{r} & =q v B \ r & =\frac{m v}{q B} \end{align*} $$

[CBSE Marking Scheme 2017]

Detailed Answer :

When a charged particle is given with initial velocity $v$ in a direction perpendicular to uniform magnetic field, it will be described in a circular path as shown. If a positively charged particle located at point $O$ is given an initial velocity $v$ perpendicular to field, the field will exert an upward force of $F=q v B$ at point $O$.

As the force is always at right angles to the velocity, it will not affect the magnitude of velocity, but will tends to change its direction. When the particle reaches points $P$ and $Q$, then in such case, directions of force and velocity will get changed. Also, a constant centripetal force will act on the particle making the particle to move in a circle.

Now the centripetal acceleration $=\frac{v^{2}}{R}$

From Newton’s second law, $q v B=\frac{m v^{2}}{R}$

Hence the radius of circular orbit will be :

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

where,

$$ m=\text { mass of particle } $$

$q=$ charge of particle