Answer:
The maximum magnetic force is 2.637 x 10⁻¹² N
Explanation:
Given;
Power, P = 8.25 m W = 8.25 x 10⁻³ W
charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C
speed of the charge, v = 314 m/s
area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²
The intensity of the radiation is calculated as;
[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]
The maximum magnetic field is calculated using the following intensity formula;
[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]
The maximum magnetic force is calculated as;
F₀ = qvB₀
F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)
F₀ = 2.637 x 10⁻¹² N