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## Homework Statement

A beam of electrons accelerated by a p.d. of 100V flies into a homogeneous magnetic field applied perpendicular to the plane of the paper towards the observer. The width of the field OP=L1=2cm. In the absence of m.f. the electron beam produces a spot at a point F on fluorescent screen AK, which is at a distance L2=6cm from teh edge of the m.f. When the m.f. is switched on the spot moves onto A along the path OQA. Find the displacement FA of the spot if the induction of the magnetic field B = 1.5 x 10

^{-3}T

## The Attempt at a Solution

Without explaining the concept- I am writing what I did in short.

Velocity gained by the electron, v= √(2eV/m)

v = 5.93 x 10

^{6}m/s

r = 0.0225m

Now for the motion of the electrons along the path OQ,

Sx = u

_{x}t

Sx = 0.02m

so t = 3.373 x 10

^{-9}s

Sy = 0.5 x v

^{2}t

^{2}/r

Sy = 8.889 x 10

^{-3}m

tanθ = Sy/Sx

θ=23.96 degrees.

This will be the angle AQE as well

so tanθ = AE/QE

from here AE = 0.0267m

So total displacement AF = AE + Sy = 0.0267 + 8.889 x 10

^{-3}

AF = 3.56cm

The answer given is 12.87cm