Question

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:

• A. \( \frac{qB}{2}[y^2 + x^2] \)
• B. \( \frac{qB y^2}{x} \)
• C. \( \frac{qB}{2}\left[\frac{y^2}{x} + x\right] \)
• D. \( \frac{qBy^2}{2x} \)

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 \(p=qBR\).

Answer 1

The Correct Option is C

Step 1: The path is a circle.
A charge $q$ entering a transverse field $B$ feels $qvB$ as a centripetal force, so it curves on a circle of radius $R$, where momentum $p = qBR$.
Step 2: Set up the geometry.
The particle enters and, after travelling so that it advances $y$ along its entry direction, is deflected sideways by $x$. The centre of the circular arc lies a distance $R$ perpendicular to the entry velocity.
Step 3: Write the circle relation.
Measuring from the centre, the entry point and the later point both lie at distance $R$. This gives \[ R^2 = y^2 + (R - x)^2. \]
Step 4: Expand and simplify.
\[ R^2 = y^2 + R^2 - 2Rx + x^2 \;\Rightarrow\; 2Rx = y^2 + x^2. \]
Step 5: Solve for the radius.
\[ R = \frac{y^2 + x^2}{2x} = \frac{1}{2}\left(\frac{y^2}{x} + x\right). \]
Step 6: Convert radius into momentum.
Using $p = qBR$, \[ p = qB\cdot\frac{1}{2}\left(\frac{y^2}{x} + x\right) = \frac{qB}{2}\left(\frac{y^2}{x} + x\right). \]
\[ \boxed{\dfrac{qB}{2}\left(\dfrac{y^2}{x} + x\right)} \]