Question

In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelerated from rest through a potential difference of 278 V. The resulting electron beam travels in a circle with a radius of 6.46 cm. The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. Assuming the magnetic field is perpendicular to the beam, find the magnitude of the magnetic field. Answer in units of T.

Answer 1

the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

Step-by-step explanation:

Given;

potential difference, V = 278 V

radius of the circular path, r = 6.46 cm = 0.0646 m

charge of electron, q = 1.60218 × 10⁻¹⁹ C

mass of electron, m = 9.10939 × 10⁻³¹ kg

The magnitude of the magnetic field is given as;

`B = (M_e*v)/(q*r)`

where;

B is the magnitude of the magnetic field

`M_e` is mass of the electron

v is velocity of the electron

r is the radius of the circular path

q is charge of the electron

Determine velocity of the electron from kinetic energy equation;

`K = (1)/(2) M_ev^2\\\\Vq = (1)/(2) M_ev^2\\\\v^2 = (2qV)/(M_e) \\\\v = \sqrt{(2qV)/(M_e)} = \sqrt{(2*1.602*10^(-19)*278)/(9.109*10^(-31))} = 9.8886*10^(6) \ m/s`

the magnitude of the magnetic field:

`B = (M_e*v)/(q*r) \\\\B = ((9.109*10^(-31))*(9.8886*10^6))/((1.602*10^(-19))*(0.0646)) = 8.704*10^(-4) \ T`

Therefore, the magnitude of the magnetic field is 8.704 x 10⁻⁴ T