An electron travels through a particular point

Question

An electron travels through a particular point in an experimental apparatus with a velocity of 0.867 × 1 0^6 m/s and at an angle of +54.7 degrees (counter-clockwise) relative to the positive x-axis in the x-y plane. If the magnetic field at this point has a magnitude of 0.01 T and is directed in along the negative y-axis, what is the magnitude of the acceleration of the electron at this point? (It is advisable to sketch a picture of the situation first.)

Answer 1

First we determine the magnetic force.

Magnetic Force

F = |q|(v)(B)(sinθ)

The magnetic force depends on the angle between the magnetic field vector and the velocity vector of the charged particle (an electron in this case). The sinθ term in the formula takes this angle into account. Note that the magnitude of the force is at a maximum when θ = ±90° and drops to zero at θ = 0° or 180°.

BTW, this is the stage where the advised sketch would be useful.

Acceleration

a = F/m = |q|(v)(B)(sinθ)/m

Acceleration is just force divided by mass of the electron.

where

|q| = 1.6 × 10^–19 C is the absolute value of the charge of an electron

v = 0.867 x 10⁶ m/s (given)

B = 0.01 T (given)

sinθ = sin(90 + 54.7) = sin(144.7) = 0.578

m = 9.11 × 10^–31 kg is the mass of an electron

As long as we are careful to express everything above in SI units, we can just plug the numbers into the formula above to calculate acceleration.

Answer

a = F/m = (1.6 × 10^–19)(0.867 x 10⁶)(0.01)(0.578)/(9.11 × 10^–31) = 8.8 × 10¹⁴ m/s²