A particle having charge 'q' enters a uniform transverse magnetic field \( \vec{B} \). It is deflected through a distance 'x' while travelling a distance 'y' as shown in figure. The magnitude of the momentum of the particle is:
Show Hint
For a charged particle moving in a magnetic field,
\[
R=\frac{p}{qB}.
\]
Whenever a deflection \(x\) and forward displacement \(y\) are given, first use circle geometry to determine \(R\), then substitute into \(p=qBR\).
Concept:
A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic force
\[
F=qvB,
\]
which acts as the centripetal force. Therefore, the particle moves along a circular path of radius
\[
R=\frac{mv}{qB}=\frac{p}{qB},
\]
where \(p\) is the momentum of the particle.
Step 1: Relate the geometrical quantities \(x\), \(y\), and \(R\).
Let the particle enter the magnetic field at point \(O\) and move along a circular arc of radius \(R\).
The centre of the circular path lies at a distance \(R\) from the point of entry. From the geometry of the figure,
\[
R^2=y^2+(R-x)^2.
\]
Expanding,
\[
R^2
=
y^2+R^2-2Rx+x^2.
\]
Cancelling \(R^2\) from both sides,
\[
2Rx=y^2+x^2.
\]
Hence,
\[
R=\frac{y^2+x^2}{2x}.
\]
Equivalently,
\[
R=\frac{y^2}{2x}+\frac{x}{2}.
\]
Step 2: Use the relation between momentum and radius.
For motion in a magnetic field,
\[
p=qBR.
\]
Substituting the value of \(R\),
\[
p
=
qB\left(\frac{y^2+x^2}{2x}\right).
\]
\[
p
=
\frac{qB}{2}
\left(
\frac{y^2}{x}+x
\right).
\]
Step 3: Identify the correct option.
Therefore, the magnitude of momentum is
\[
\boxed{
p=
\frac{qB}{2}
\left(
\frac{y^2}{x}+x
\right)
}
\]
which corresponds to option \((C)\).
