Question

Thanks in advance, I couldn't find the answers to the following questions:

Q1) A particle (*q* = –4.0 *µ*C, *m* = 5.0
mg) moves in a uniform magnetic field with a velocity having a
magnitude of 2.0 km/s and a direction that is 50° away from that of
the magnetic field. The particle is observed to have a magnetic
force with a magnitude of 29 ?N. What is the magnitude of the
magnetic field?

Q2) An electron has a velocity of 6.0 × 10^{6} m/s in
the positive x-direction at a point where the magnetic field has
the components *B**x* = 3.0 T, *B**y* =
1.5 T and *B**z* = 2.0 T. What is the magnitude of
the magnetic force on the electron at this point?

Q3) A proton has a speed 6.9 × 10^{5} m/s enters
perpendicularly into a uniform 0.50 T magnetic field region. What
is the radius of the resulting path? (proton: *m* = 1.67 ×
10^{?27}kg, *q* = 1.6 × 10^{?19} C)

Answer #1

1) The magnetic force on the particle is

where theta is the angle between the charge and magnetic field

I am assuming F is 29 N

q = -4 µC= -4 x 10^{-6} C

v = 2 km/s = 2000 m/s

Therefore the magnitude of the magnetic field is

2) The magnetic force is

since v is in x direction, the y component of the force is

The z component of the force is

The magnitude of the net force is

3) For the proton to move in a circular path in the magnetic field, its magnetic force must balance the centripetal force

Therefore the radius of the resulting path is

a) An electron has a velocity of 7.0 x 106 m/s in the
positive x direction at a point where the magnetic field has the
components Bx = 3.0 T, By = 2.5 T, and Bz =
2.0 T. What is the magnitude of the acceleration of the electron at
this point (in 1018 m/s2)? (me = 9.11 x
10-31 kg, e = 1.6 x 10-19 C)
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(1 )An alpha particle (q = +2e, m =
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A particle with charge q = 6.0 nC and mass
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A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with
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The magnetic field extends for a distance D = 0.75 m in the
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magnitude E=60.0 N/C. At time t = 0 the
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A particle (q = -4.0 μC, m = 5.0 mg) moves in a uniform magnetic
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