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;
[tex]B = \frac{M_e*v}{q*r}[/tex]
where;
B is the magnitude of the magnetic field
[tex]M_e[/tex] 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;
[tex]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[/tex]
the magnitude of the magnetic field:
[tex]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[/tex]
Therefore, the magnitude of the magnetic field is 8.704 x 10⁻⁴ T
