5) an electron gun sends out a beam of electrons whose kinetic energies are all about 57 μμev. (1 μev=1.6×10−25 j1 μev=1.6×10−25 j.) you need to set up a magnetic field perpendicular to the beam that causes it to turn through a 90∘∘ circular arc of length 2 mm. how strong must the magnetic field be? [note: electrons have a mass of about 9.11×10-31 kg.]

2 x 10^-5 Tesla

Explanation:

Kinetic energy, K = 57 micro electron volt = 57 x 1.6 x 10^-25 J

Length of arc = 2 mm

Angle turn = 90° = π/2 = 1.57 radian

Let the strength of magnetic field is B.

mass of electron, m = 9.11 x 10^-31 kg

K = 1/2 x mv²

where, v be the velocity of electron

57 x 1.6 x 10^-25 = 0.5 x 9.11 x 10^-31 x v²

v = 4474.59 m/s

radius = arc length / angle = 2 / 1.57 = 1.27 mm

By the formula for the radius

r = mv / Bq

B = mv / qr

B = (9.11 x 10^-31 x 4474.59) / (1.6 x 10^-19 x 1.27 x 10^-3)

B = 2 x 10^-5 tesla

Thus, the magnetic field is 2 x 10^-5 Tesla.

i think the answer would be reserve.