Please help, I can\'t figure out how to go about this one, fromSchaums - answer

Question

Please help, I can't figure out how to go about this one, fromSchaums - answer should be 5.2 cm.
Q: An electron is accelerated from rest through apotential difference of 3750 V. It enters a region whereB = 4.0 x 10-3 T perpendicular to itsvelocity. Calculate the radius of the path it willfollow.
Thanks.
Q: An electron is accelerated from rest through apotential difference of 3750 V. It enters a region whereB = 4.0 x 10-3 T perpendicular to itsvelocity. Calculate the radius of the path it willfollow.
Thanks.

Explanation / Answer

As the electron enters themagnetic field, it has an energy
of 3750 eV.

So its KE is 0.5mv² = 3750 eV...(i)

where m = 9.109 × 10^(-31) kg is the electron mass, and v isits speed

1 eV = 1.6 * 10^(-19) J

So 0.5mv² = 3750 * 1.6 * 10^(-19) J

0.5mv² = 6.000 * 10^(-16) J

And v = [2 * 6.000* 10^(-16) / {9.109 × 10^(-31)}J/kg]

giving v = 3.6296 * 10^7 m/s....(ii)

The magnetic force = the centrifugal force in the magneticfield:

qvB = mv²/r

where q = 1.602 × 10^(-19) coulombs (C) is the electroncharge,
v = electron speed, m its mass, B = 4.0 mT is the magnetic fieldstrength
and r is the radius of the circular path.

So qB = mv/r

and r = mv/(qB)

r = {9.109 × 10^(-31) kg * 3.6296 * 10^7 m/s} / {1.602× 10^(-19) C * 4.0 * 10^(-3)T}

r = 5.1595 * 10^(-2) m

or r = 5.160 cm

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