10 5 Velocity Selector – Class 11TH Physics

10.5 Velocity Selector

A velocity selector is a device used to determine the velocity of a charged particle. In this device, electric and magnetic forces are applied to the moving particle in such a way that they balance each other only for one value of velocity, allowing the particle to continue moving with a constant velocity.

Consider a particle with a positive charge $+q$ that enters a uniform magnetic field $B$ at a right angles to it, with a velocity $v$ . The magnetic force acts on the particle in the upward direction, as shown in Fig. 10.11. To balance this magnetic force, an electric force must act downward on the particle.

A velocity selector consists of a cylindrical tube located within a magnetic field $B$ . Inside the tube is a parallel plate capacitor that creates a uniform electric field $E$ . The electric field $E$ is oriented perpendicular to the magnetic field $B$ , as shown in Fig. 10.12.

When the charged particle enters the left end of the tube, the magnetic force acts upward, while the electric force acts downward in the direction of the electric field $E$ on the particle.

The magnitude of the velocity selected can be determined as below.

As the velocity $v$ is perpendicular to both $B$ and $E$ , therefore:

Magnetic force (upward) $= Bqv$
Electric force (downward) $= qE$

For no deflection of particle: $Bqv = qE$

v = \frac{E}{B} \quad \ldots (10.15)

Example 10.4

Alpha particles ranging in speed from $1000\,\mathrm{m\,s^{-1}}$ to $2000\,\mathrm{m\,s^{-1}}$ enter into a velocity selector where the electric intensity is $300\,\mathrm{V\,m^{-1}}$ and the magnetic induction $0.20\,\mathrm{T}$ . Which particle will move undeviated through the field?

Solution

$E = 300\,\mathrm{V\,m^{-1}} = 300\,\mathrm{N\,C^{-1}}$ , $B = 0.20\,\mathrm{T}$

Only those particles will be able to pass through the plate for which the electric force $qE$ acting on the particles balances the magnetic force $Bqv$ on the particles as shown in the Fig. 10.12.

Therefore $qE = Bqv$

Thus, the selected speed is: $v = E/B = 300\,\mathrm{N\,C^{-1}} / 0.20\,\mathrm{N\,A^{-1}\,m^{-1}} = 1500\,\mathrm{m\,s^{-1}}$

Point to ponder!

lightbulb

A force is exerted on a moving charged particle in a magnetic field. In what direction it should move so that the force is not exerted on it?

Example 10.5

A charged particle moves through a velocity selector at a constant velocity in a straight line. The electric field of the velocity selector is $4.8 \times 10^{3}\,\mathrm{N\,C^{-1}}$ , while the magnetic field is $0.2\,\mathrm{T}$ . When the electric field is turned OFF, the charged particle travels on a circular path of radius $3.0\,\mathrm{cm}$ . Find the charge to mass ratio of the particle.

Solution

Since the particle is moving in a direction perpendicular to both $E$ and $B$ , so the magnitude of velocity $v$ will be given by $qE = Bqv$ or $v = E/B$

When the electric field is turned off, the particle will move along a circular path of radius $r$ . Then the magnetic force provides necessary centripetal force. Then:

Bqv = \frac{mv^{2}}{r}

or $q/m = v/(Br)$

Putting the values of $E$ , $B$ and $r$ , we have:

\frac{q}{m} = \frac{E/B}{Br} = \frac{E}{B^{2}r}

= \frac{4.8 \times 10^{3}\,\mathrm{N\,C^{-1}}}{(0.20\,\mathrm{T})^{2}(3 \times 10^{-2}\,\mathrm{m})}

= \frac{4.8 \times 10^{3}}{1.2 \times 10^{-3}}

= 4 \times 10^{6}\,\mathrm{C\,kg^{-1}}