An electron is accelerated through 1.95 103 V from rest and then enters a uniform 1.50-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience?

As stated by Lorentz Force law, the magnitude of a magnetic force, F, can be expressed in terms of a fixed amount of charge, q, which is moving at a constant velocity, v, in a uniform magnetic field, B, as follows;

F = qvB sin θ ------------(i)

Where;

θ = angle between the velocity and the magnetic field vectors

When the electron passes through a potential difference, V, it is made to accelerate as it gains some potential energy ([tex]P_{E}[/tex]) which is then converted to kinetic energy ([tex]K_{E}[/tex]) as it moves. i.e

[tex]P_{E}[/tex] = [tex]K_{E}[/tex] ----------------(ii)

But;

[tex]P_{E}[/tex] = qV

And;

[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex] x m x v²

Therefore substitute these into equation (ii) as follows;

qV = [tex]\frac{1}{2}[/tex] x m x v²

Make v subject of the formula;

2qV = mv²

v² = [tex]\frac{2qV}{m}[/tex]

v = [tex]\sqrt{\frac{2qV}{m} }[/tex] ---------------(iii)

From the question;

q = 1.6 x 10⁻¹⁹C (charge on an electron)

V = 1.95 x 10³V

m = 9.1 x 10⁻³¹kg

Substitute these values into equation (iii) as follows;

v = [tex]\sqrt{\frac{2*1.6*10^{-19} * 1.95*10^{3}}{9.1*10^{-31}} }[/tex]

v = [tex]\sqrt{\frac{6.24*10^{-16}}{9.1*10^{-31}} }[/tex]

v = 2.63 x 10⁷m/s

Now, from equation (i), the magnitude of the magnetic force will be maximum when the angle between the velocity and the magnetic field is 90°. i.e when θ = 90°

Substitute the values of θ, q v and B = 1.50T into equation (i) as follows;

F = qvB sin θ

F = 1.6 x 10⁻¹⁹ x 2.63 x 10⁷ x 1.50 x sin 90°

F = 6.3 x 10⁻¹²N

Therefore the maximum magnitude of the magnetic force this particle can experience is 6.3 x 10⁻¹²N