Question

A capacitor is connected to a battery and placed in a magnetic field (-z direction) to form a velocity selector. A charge moving at 3,479.6 m/s is not deflected by the velocity selector. If the voltage of the battery is 7 volts and the distance between the sheets is 0.3 meters, what is the amount of the magnetic field in milli-Tesla?

Answer 1

We know that velocity selector is a region in which there is uniform electric and mgnetic field which are perpendicular to each other and also perpendicular to the initial velocity of the charged particle.

Now, we know that force on a charged particle due to the mgnetic field , FM  is given by:

FM = q( v*B) ( q = magnitude of charge

v = velocity of the charged particle

B = applied Magnetic field )

where v*B = vBSin(theta)

theta is the angle between velocity of charged particle and the applied magnetic field.

Since v is perpendicular to B as mentioned above so that theta = 90 degree , hence

v*B = vBSin 90 = vB ( Sin 90=1)

so that   FM = qvB

  

Also, the force due to electric field on the charged particle, FE is given by :

FE = qE ( E = Applied Electric Field)

In velocity selector, force on a charged particle due to electric field ( FE ) would counteract the force due to magnetic field (FM.) i.e

FM = FE

qvB = qE ........ eqn 1

By cancelling out q on both sides of equation 1, we get

vB = E ..........eqn 2

or v= E/B

or B = E/v    .......... eqn 3

Now, velocity of charged particle, v = 3479.6 m/s

Applied voltage, V = 7 volts

Distance between the sheets , d = 0.3 m

We know that,

E = V/d .............. eqn 4

Using the values of V and d in eqn 4,  we get

E = 7/ 0.3 = 23.333 Vm-1

Since we are to calculate the applied Magnetic field, hence we would put the value of v and E in eqn 3

i.e B = E/v = 23.333/ 3479.6 = 0.006705 T = 6.70* 10-3 T = 6.7 mT (Because 1 mT = 10-3 T

6.7mT = 6.7* 10-3 T )

Hence, the magnetic field B has a magnitude = 6.7 mT