An electron in the beam of a tv picture tube is accelerated through a potential difference of 2.00 kv.? it then passes into a magnetic field perpendicular to its path, where it moves in a circular arc of diameter 0.360m. find the magnitude of the field?

The magnitude of the field is 8.384×10^-4 T.

Explanation:

Now, i start solving this question:

First, convert the potential difference(V) 2 kv to 2000 v.

As, we have the final formula is qvB = mv^2/r. It came from the centripetal force and the magnetic force and we know that these two forces are equal. When dealing with centripetal motion use the radius and not the diameter so

r = 0.36/2 = 0.18 m.

As, we are dealing with an electron so we know its mass is 9.11*10^-31 kg and its charge (q) is 1.6*10^-18 C.

We can solve for its electric potential energy by using ΔU = qV and we know potential energy initial is equal to kinetic energy final so ΔU = ΔKE and kinetic energy is equal to 1/2mv^2 J.

qV = 1/2mv^2

(1.6*10^-19C)(2000V) = (1/2)(9.11*10^-31kg) v^2

v = 2.65×10^7 m/s.

These all above steps we have done only for velocity(v) because in the final formula we have 'v' in it. So, now we substitute the all values in that formula and will find out the magnitude of the field:

qvB = mv^2/r

qB = mv/r

B = mv/qr

B = (9.11*10^-31 kg)(2.65×10^7 m/s) / (1.6*10^-19 C)(0.18 m)

Hence, B = 8.384*10^-4 T.