1. One component of a magnetic field has a magnitude of 0.028 T and points along
ID: 1282647 • Letter: 1
Question
1. One component of a magnetic field has a magnitude of 0.028 T and points along to +x axis, while the other component has a magnitude of 0.078 T and points along the -y axis. A particle carrying a charge of +1.50 x 10^-5 C is moving along the +z axis at a speed of 2.20 x 10^3 m/s.
a) find the magnitude of the net magnetic force that acts on the particle? answer in N
b) determine the angle that the net force makes with respect to the +x axis?
2. A charged particle moves through a velocity selector at a contant speed in a straight line. The electric field of the velocity selector is 4.35 x 10^3 N/C, while the magnetic field is 0.360 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.95 cm. Find the charge-to-mass ratio of the particle.
answer in C/kg
3. The drawing shows a thin, uniform rod that has a length of 0.50 m and a mass of 0.080 kg. This rod lies in the plane of the screen and is attached to the floor by a hinge at point P. A uniform magnetic field of 0.39 T is directed perpendicularly into the plane of the screen. There is a current I = 3.8 A in the rod, which does not rotate clockwise or counter-clockwise. Find the angle theta. (hint: magnetic force may be taken to act at the center of gravity.)
Explanation / Answer
Q = 1.5 * 10-5 C, v = (2.2 * 103 m/s)k
B = (0.028 T)i+ ( 0.078 T)(-j)
F = Q (v x B)
F = (1.5 * 10-5)((2.2 * 103)kx (0.028T)i+ ( 0.078 T)(-j))
F = (9.24 * 10-4)j + (2.574 * 10-3 i) N
magnitude F = 2.73 *10-3 N
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b) theta = tan-1 Fy/Fx
=tan-1( 9.24 * 10-4 N/ 2.574 *10-3N)
= 19.75 deg
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2) FB = FE
q v B = q E
r = m v / q B
q/m = E / r B2 = (4.35 x 103) / (4.95 * 10-2 * 0.3602)
= 6.8 * 105 C/kg
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3) l = 0.50 m, B = 0.39 T, i = 3.8 A, m = 0.080 kg
Magnetic Force F = i l B
mg = i l B
m g * l / 2 cos theta = i l B (l / 2 )
cos theta = i B l / mg
= (3 .8 A * 0.39 T * 0.50 m) / (0.080 kg * 9.8m/s2)
= cos-1 ( 0.95 )
theta = 19.10
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