Question

An electron is accelerated through 2.70 103 V from rest and then enters a uniform 1.40-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience?

Answer 1

To solve the exercise it is necessary to apply the concepts of Mangenetic Force and Energy generated by electric potential.

By definition we know that the energy within an electron through voltage can be expressed as

$E= eV$

Where,

e= charge of electron

V= Voltage

The kinetic energy of a moving system can be expressed as

$\Delta KE = \frac{1}{2}mv^2$

Where,

m = mass

v = Velocity

For energy conservation we have to

$E= \Delta KE$

$eV = \frac{1}{2}mv^2$

Solving to find v,

$v= \sqrt{\frac{2eV}{m}}$

On the other hand we have that the Magnetic Force can be expressed as,

$F=qvBsin\theta$

Where,

q=charge of proton

v=velocity

B= Magnetic field

$\theta =$ Angle between the magnetic field and the velocity vector (It is perpendicular in this case)

Using the previous equation from velocity in the Force equation we have,

$F=qBsin\theta(\sqrt{\frac{2eV}{m}})$

$F=eBsin\theta(\sqrt{\frac{2eV}{m}})$

PART A) Replacing the values to find the force we have,

$F_{max} = eB\sqrt{\frac{2eV}{m}}$

$F_{max} = (1.6*10^{-19})(1.4)sin(90)(\sqrt{\frac{2(1.6*10^{-19})(2.7*10^3)}{9.11*10^{-31}}})$

$F_{max} = 6.8983*10^{-12}N$