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Physics Courseware
Physics II
Magnetic Force
Force on a charge moving in a magnetic field:
Force on a wire carrying a current I:
F = BqvSin∠ Bv
F = BIlSin∠lB
Problem 1: Find the magnitude and indicate the direction of the force on a charge of 6.25
µC for each diagram shown, if B=1.5 T and v=29,979 m/s:
Solution: The magnitude of the force is given by F=BqvSin(θ), so for cases (a) and (b)
the force will be F=(1.5 T)( 6.25 µC)( 29,979 m/s)= 0.28N. The force in case (c) is zero
because the magnetic field and the velocity are collinear.
We find the direction of the force in cases (a) and (b) using the right hand rule:
Problem 1a: Find the magnitude and direction of the force on a positive charge of 2.5
µC for each diagram shown, if B=0.8 T and v=1,200 m/s:
Solution: The magnitude is F = BqvSin∠ Bv = 0.8(2.5 × 10-6 )(1200)Sin90o = 0.0024 N
Problem 2: If two adjacent parallel wires carry electric current in the same direction, are
they attracted to each other or repelled from each other? Give a short explanation.
Solution: If two adjacent parallel wires carry electric current in the same direction, they
will attract each other. To prove this consider one wire first and determine the magnetic
field created at the position of the second wire and then use the right hand rule to
determine the direction of the force on the second wire.
Problem 3: The particle shown in the figure enters a region of magnetic field and is
deflected upward. Is the charge of the particle positive or negative? Explain.
Solution: The right hand rule indicates that the particle is positive.
Problem 4.- Find the magnetic force on a proton coming towards our planet at a speed of
4x105m/s if the magnetic field of the Earth is 1.5x10-5T at that point and it is
perpendicular to the velocity of the proton (as shown in the figure). Answer with
magnitude and indicate the direction of the force in the figure below.
Solution: The force on a positive charge with the velocity and magnetic field vectors as
shown in the figure can be found using the FBI rule. The force vector is shown in red:
The magnitude of the force is given by:
F = BqvSinθ = (1.5 × 10 -5 T)(1.6 × 10 -19 C)(4 × 105 m/s) = 9.6x10-19N
Problem 5.-Two parallel wires carry current in the same direction. Is the magnetic force
repulsive or attractive?
Solution: The magnetic force is attractive in this case.
Problem 6.- A 2-meter long wire has a mass of 0.0035 kg and is in a place where the
magnetic field is horizontal, perpendicular to the wire and has a magnitude B=0.9 tesla.
Calculate the current needed so the wire will “float” due to the magnetic force.
Solution: For the wire to float its weight will have to be equal to the magnetic force:
We use the equation F = BIl sin ∠lB Force on a current carrying wire in a magnetic field:
mg
0.0035 × 9.8
mg = BIl sin ∠lB → I =
=
= 0.019 A
l
Bl sin ∠ B 0.9 × 2 × sin 90o
This is another one-liner. Three points for the right equation, full credit for the correct
answer.
Problem 7.- The following charged particles are moving in the proximity of a current
carrying wire. The sign of the charges is indicated (+ or -) as well as their velocity (with
arrows).
For each charge indicate the direction of the magnetic force due to the magnetic field
produced by the wire.
Solution: The current generates a magnetic field given by the right hand rule which
points in the directions indicated by the red crosses and dots.
Based on those directions we use the left hand rule to find the forces on the charges
(forces shown as green arrows)
Problem 8.- Determine the magnitude and direction of the force on an electron
traveling at a speed of 5.75x105 m/s horizontally to the east in a the presence of a
vertically upward magnetic field of 0.85T
[charge of the electron qe= -1.6x10-19 C]
Solution: The “FBI” rule tells us the direction of the force, but notice that because the
electron has negative charge we have to reverse the direction given by the rule. As shown
in the figure, the force will be due North.
The magnitude of the force is determined by:
F=qvBsin(θ)=(1.6x10-19 C)( 5.75x105 m/s)( 0.85T)(1)= 7.82x10-14 N
Problem 9: Calculate the magnitude of the force on one meter of wire due to the other.
Take the currents to be 25A and the distance between the wires 0.8m
Solution: The magnitude of the force on one meter of wire due to the other is:
4π × 10 −7 (25)
µo I
l
l
F = BIlSin∠ B =
IlSin∠ B =
(25)Sin90o = 1.56x10-4 N
2πr
2π (0.8)
Problem 10.- Describe the trajectory of an electron that enters a region of constant
magnetic field at right angles (the angle between the field and the velocity is 90º).
Solution: The trajectory of an electron (or any other charged particle) in a region of
constant magnetic field entered at right angles is a circle (recall the example of the mass
spectrometer).
Problem 11.- Determine the radius of the circular motion of a carbon-12 ion in a region
of magnetic field B=0.85T if its speed is 5.75x106 m/s perpendicular to the magnetic
field.
[charge of the carbon-12 ion = 1.6x10-19 C, mass of the ion=2.0x10-26kg