Chapter 17
Problem 1
The magnetic field always points to the floor. For (a), point your right hand infront of you, then curl it
towards the ground. Note that your thumb points to your left. If you are facing north, the directino to
your left is west and your right is east. However, this is an electron, so we must flip the direction of your
thumb giving east. For (b), Keep your hand in the same position as (a), but now rotate 30 degrees to the
west (left). Notice your thumb moves by the same amount, giving 30 degrees north of east. For (c), the
component pointing in the downward direction is parallel to the field, so it doesn’t matter. Point your fingers
northward, and curl down, again giving east. For (d), the field and velocity are parallel, so the sin(theta)
term gives zero.
Problem 3
Just use the right hand rule. Thumb points to the force, fingers point towards velocity. The direction your
fingers curl is the field. (a) Into page. (b) To the right. (c) Down.
Problem 11
The current is flowing to the right, and the force points down, using the same procedure as in problem 3, this
gives a magnetic field pointing out of the page. It’s magnitude is F/l = I ∗ B → B = Fl · I1 = (0.12)(1/15)T
Problem 15
F = IB` sin θ = 10 · 0.3 · 5 sin(30) = 125N
Problem 27
qBr
Using r = mv
qB , the velocity is v = m . The velocity is also equal to distance over time, v =
two equations equal to each other and solve for B. Notice that the r cancels from both sides.
2πr
t .
Set these
Problem 31
Again we use r = mv
qB . We know everything except the velocity which we can figure out. The particle starts at
rest and goes through a potential difference of 250V . In doing so, it gains kinetic energy (it was accelerating)
in the amount 12 mv 2 = q∆V = 4 · 10−17 J. Multiply by two, divide by the mass and solve for v. Take this
velocity and plug it into the equation for the radius.
Problem 35
The magnetic field from a wire is given by B =
µ0 I
2πR .
The distance is 100m and the current is given as 104 A.
Problem 37
Using the same expression as above, we are looking for the distance from the wire at which the magnetic
field is 1.7mT . Plug in I = 20A and B = 1.7 · 10−3 T and solve for the radius.
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