An electron moves at a speed of 5.2 multiply.gif 106 m/s perpendicular to a cons
ID: 1769219 • Letter: A
Question
An electron moves at a speed of 5.2 multiply.gif 106 m/s perpendicular to a constant magnetic field. The path is a circle of radius 1.9 multiply.gif 10-3 m.
(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 singly ionized 24Mg ion (mass 3.983 multiply.gif 10-26 kg) is accelerated through a 2.6 kV potential difference and deflected in a magnetic field of 560 G in a mass spectrometer.
(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
A metal bar is pulled along a U-shaped wire as shown in the figure. The external magnetic field has a magnitude of 30 T. The velocity of the wire is constant at 39 m/s. The length, script lowercase l, of the wire is 0.2 m and the initial position, x, of the bar is 1.2 m. The resistance of the bar is 26 capital omega and that of the wire is negligible. What is the current through the wire and bar?
_________________ A
Explanation / Answer
1)
a) a = v^2/r
= (5.2*10^6)^2/(1.9*10^-3)
= 1.423*10^16 m/s^2 <<<<<<<----------Answer
b) m*v^2/r = q*v*B
B = m*v/(r*q)
= 9.1*10^-31*5.2*10^6/(1.9*10^-3*1.6*10^-19)
= 1.56*10^-2 T <<<<<<<----------Answer
2)
a)
we know, 1G = 10^-4 T
B = 0.26*10^-4 T
E = 490 v/m
for not to defelected
Fe = Fb
q*E = q*v*B
==> v = E/B
= 490/0.26*10^-4
= 1.884*10^7 m/s <<<<<<<----------Answer
b)
KE = 0.5*m*v^2
= 0.5*1.67*10^-27*(1.884*10^7)^2
= 2.964*10^-13 J
= 2.964*10^-13/1.6*10^-19 eV
= 1.85*10^6 eV
= 1850 keV <<<<<<<----------Answer
c)
KE = 0.5*m*v^2
= 0.5*9.1*10^-31*(1.884*10^7)^2
= 1.615*10^-16 J
= 1.615*10^-16/1.6*10^-19 eV
= 1009.4 eV <<<<<<<----------Answer
3)
a)
gain in KE = Workdone
0.5*m*v^2 = q*V
v = sqrt(2*q*V/m)
= sqrt(2*1.6*10^-19*2600/3.983*10^-26)
= 1.67*10^5 m/s
m*v^2/r = q*v*B
r = m*v/(B*q)
= 3.983*10^-26*1.67*10^5/(560*10^-4*1.6*10^-19)
= 74 m <<<<<<<----------Answer
b) r2 - r1 = 26*74/24 - 74
= 6.17 m <<<<<<<----------Answer
4)
a) flux = N*A*B*cos(90)
= 0 <<<<<<<----------Answer
b) flux = N*A*B*cos(0)
= 50*pi*0.06^2*0.7*10^-4*1
= 3.97*10^-5 wb <<<<<<<----------Answer
c) flux = N*A*B*cos(90)
= 0 <<<<<<<----------Answer
d)flux = N*A*B*cos(30)
= 50*pi*0.06^2*0.7*10^-4*cos(30)
= 3.43*10^-5 wb <<<<<<<----------Answer
5) imduced emf = B*v*L*sin(90)
= 30*39*0.2*1
= 234 volts
I = emf/R
= 234/26
= 9 A <<<<<<<----------Answer
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