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Chapter 26: Problem 47

Show that the orbital radius of a charged particle moving at right angles to a magnetic field \(B\) can be written \(r=\sqrt{2 K m / q B}\) where \(K\) is the kinetic energy in joules, \(m\) the particle's mass, and \(q\) its charge.

Short Answer

Expert verified
The orbital radius \(r\) of a charged particle moving at right angles to a magnetic field \(B\) is \(r=\sqrt{2 K m / q B}\) when the magnetic force on the particle provides the necessary centripetal force and the particle's velocity is given in terms of its kinetic energy.

Step by step solution

01

Define the Magnetic Force

The magnetic force \(F\) acting on a moving charged particle can be given by \(F = qvB\), where \(v\) is the velocity of the particle, \(B\) is the magnetic field, and \(q\) is the charge of the particle.
02

Define the Centripetal Force

The centripetal force \(F_c\) needed to keep a particle moving in a circular path is given by \(F_c = mv^2/r\), where \(m\) is the mass of the particle, \(v\) is the velocity of the particle and \(r\) is the radius of the circular path.
03

Equating Magnetic Force to Centripetal Force

Since the magnetic force provides the necessary centripetal force, equate the two equations from step one and two: \(qvB = mv^2/r\). Simplify the equation to obtain the relation for \(r\), \(r = mv/qB\).
04

Substitute the Kinetic Energy

The kinetic energy \(K\) of a particle is given by \(K = 1/2 mv^2\). Solve this equation for \(v\) to get \(v = \sqrt{2K/m}\) and substitute this obtained expression for \(v\) in the equation from step 3 to get, \(r = \sqrt{2 K m / q B}\).

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