Respuesta :
Answer:
Part a)
force will be along East direction
Part b)
[tex]a = 6.27 \times 10^{14} m/s^2[/tex]
Part C)
[tex]x = 2.96 \times 10^{-3} m[/tex]
Explanation:
As we know that vertical component of the magnetic field is given as
[tex]B_y = 55 \mu T[/tex]
Kinetic energy of the electron beam is given as
[tex]K = 12 keV[/tex]
Part a)
As we know that force on moving charge due to magnetic field is given as
[tex]\vec F = q(\vec v \times \vec B)[/tex]
now we know that
[tex]\vec F = (-e)(v \hat j \times B (-\hat k))[/tex]
[tex]\vec F = evB \hat i[/tex]
so force will be along East direction
Part b)
As we know that
[tex]K = \frac{1}{2}mv^2[/tex]
[tex]\frac{1}{2}mv^2 = 12 \times 10^3 (1.6 \times 10^{-19}) J[/tex]
[tex]v = 6.5 \times 10^7 m/s[/tex]
now acceleration of the charge is given as
[tex]a = \frac{evB}{m}[/tex]
[tex]a = \frac{(1.6 \times 10^{-19})(6.5 \times 10^7)(55 \times 10^{-6})}{9.11 \times 10^{-31}}[/tex]
[tex]a = 6.27 \times 10^{14} m/s^2[/tex]
Part C)
times taken by the beam of electron to move through the distance of 20 cm is given as
[tex]t = \frac{L}{v}[/tex]
[tex]t = \frac{0.20}{6.5 \times 10^7}[/tex]
[tex]t = 3.08 \times 10^{-9} s[/tex]
so the deflection in the direction of magnetic field is given as
[tex]x = \frac{1}{2}at^2[/tex]
[tex]x = \frac{1}{2}(6.27 \times 10^{14})(3.08 \times 10^{-9})^2[/tex]
[tex]x = 2.96 \times 10^{-3} m[/tex]
