(a) Find the magnitude of the electron\'s acceleration. ______________________ m
ID: 1769213 • Letter: #
Question
(a) Find the magnitude of the electron's acceleration.
______________________ m/s2
(b) What is the magnitude of the field?
_____________________ T
A velocity selector has a magnetic field of magnitude 0.26 G perpendicular to an electric field of magnitude 0.49 kV/m.
(a) What must the speed of a particle be for it to pass through undeflected?
____________________ m/s
(b) What energy must protons have to pass through undeflected?
_________________ keV
(Note: 1 eV [electron volt] = 1.602 x 10-19 J.)
(c) What energy must electrons have to pass through undeflected?
________________ eV
(a) Find the radius of curvature of the orbit for the ion.
_________________ cm
(b) What is the difference in radius for 26Mg and 24Mg ions? (Assume that their mass ratio is 26/24.)
____________________ cm
A circular coil has 50 turns and a radius of 6 cm. It is at the equator, where the earth's magnetic field is 0.7 G north. Find the magnetic flux through the coil when its plane is as follows.
(a) horizontal
___________________ Wb
(b) vertical with its axis pointing north
___________________ Wb
(c) vertical with its axis pointing east
____________________ Wb
(d) vertical with its axis making an angle of 30 with north
___________________ Wb
Explanation / Answer
1)
The magnetic force on the proton is:
Fm = qv x B = q v x^ x By^ = (qvB) x^ x y^ = (qvB) z^
q = elementary unit of charge = 1.6e-19 (C)
B = 3.3 e-5 (T)
The gravitational force on the proton is:
Fg = -mg z^
m = proton mass = 1.67e-27 (kg)
g = local acceleration of gravity = 9.8 (m/s^2)
Therefore, if the forces balance:
qvB z^ = Fm = - Fg = mg z^
or
qvB = mg
or
v = mg/(qB) = (1.67e-27)(9.8)/((1.6e-19)(3.3e-5))
= ( (1.67 * 9.8)/(1.6*3.3) )e-3 = 3.1e-3 (m/s)
So v = 3.1*10^(-3) (m/s)
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