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

Answer:

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

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;

$B = \frac{M_e*v}{q*r}$

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;

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

the magnitude of the magnetic field:

$B = \frac{M_e*v}{q*r} \\\\B = \frac{(9.109*10^{-31})*(9.8886*10^6)}{(1.602*10^{-19})*(0.0646)} = 8.704*10^{-4} \ T$

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