Question: Q. 2. A particle of charge $q$ is moving with velocity $v$ in the presence of crossed electric field $E$ and magnetic field $B$ as shown. Write the condition under which, the particle will continue moving along $x$-axis. How would the trajectory of the particle be affected if the electric field is switched off ?
R] [SQP 2017-18]
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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