Question

A particle of charge $ q $, mass $ m $, and kinetic energy $ E $ enters in a magnetic field perpendicular to its velocity and undergoes a circular arc of radius $ r $. Which of the following curves represents the variation of $ r $ with $ E $?

• A.
• B.
• C.
• D.

Hint:

For a charged particle moving in a magnetic field, the radius of the circular path is proportional to the square root of its kinetic energy.

Answer 1

The Correct Option is D

Given: - The particle has charge \( q \), mass \( m \), and kinetic energy \( E \). - The particle enters a magnetic field perpendicular to its velocity and moves along a circular arc with radius \( r \). The radius \( r \) for a charged particle moving in a magnetic field is given by: \[ r = \frac{mv}{qB} \] The kinetic energy of the particle is \( E = \frac{1}{2} mv^2 \). Solving for \( v \): \[ v = \sqrt{\frac{2E}{m}} \] Substituting the value of \( v \) into the formula for \( r \): \[ r = \frac{m\sqrt{\frac{2E}{m}}}{qB} = \frac{\sqrt{2mE}}{qB} \] 
Thus, the radius \( r \) is proportional to the square root of the kinetic energy \( E \): \[ r \propto \sqrt{E} \] The relationship between \( r \) and \( E \) is represented by a curved relationship, as shown in option (4). 
Final Answer (4)

Answer 2

Given the equation for the motion of a charged particle in a magnetic field: \[ \frac{mv^2}{r} = qvB \] This simplifies to: \[ mv = qBr \] The energy of the particle is: \[ E = \frac{1}{2} mv^2 \] Substituting for \(mv\): \[ E = \frac{1}{2} m \left( \frac{q^2 B^2 r^2}{m^2} \right) = \frac{q^2 B^2 r^2}{2m} \] Thus, we get: \[ E = \left( \frac{q^2 B^2}{2m} \right) r^2 \] This shows that: \[ r^2 \propto E \] And the graph of \(r\) vs. \(E\) is shown as: \[ \boxed{r^2 \propto E} \]