The Earth has a magnetic field of its own. The north pole of a magnet is attracted to the North magnetic pole of the Earth, and the south pole to the South magnetic pole, hence the choice of north and south for designating the poles of a magnet.
Since the North magnetic pole attracts the north pole of a magnet, the North magnetic pole of the Earth is a south pole. And likewise, the South magnetic pole of the Earth is a north pole. In fact, the magnetic field of the Earth is approximately the field of a large bar magnet within the Earth, with the south pole of the bar magnet near the North magnetic pole and the north pole of the magnet near the South magnetic pole.
The North and South magnetic poles aren't located at the North and South poles of the Earth. (Use globe.) The North magnetic pole is somewhat north of Hudson's Bay in Canada, about 1300 miles from the North pole.
A charged particle moving through a magnetic field feels a force given by
The expression for the force, F=qvBsinq, is greatest when sinq = 1, or q = 90°. This corresponds to the charge moving perpendicular to the magnetic field, B. If the charge moves parallel to the magnetic field, the force is zero.
The direction of the force is perpendicular to both the velocity and the magnetic field. Use the right hand rule (version #1) to determine the direction of the force:
Hold your right hand open, and orient your fingers in the direction of B (follow the direction of the arrows if available) and point your thumb in the direction of v (this may require you to rotate your hand). Now your palm is facing the direction of the force, F, exerted on a positive charge.
An electron gun fires electrons into a magnetic field that is directed straight downward. find the direction of the force exerted on an electron by the field for each of the following directions of the electron's velocity: (a) horizontal and due north; (b) horizontal and 30° west of north; (c) due north, but at 30° below the horizontal; (d) straight upward. (Remember that an electron has a negative charge.)
Make a sketch of this scenario showing downward magnetic field, and an axis pointing north. To be sure you can get everything, you may want to draw both a front view and a side view.
(a) Use the right hand rule, fingers pointing down in the direction of B, thumb pointing north. Your palm faces west. Since this is a negative charge, the force is directed oppositely, that is to the east.
(b) Fingers pointing down, thumb pointing 30° west of north. Your palm faces 30° south of west, so the force is 30° north of east.
(c) Fingers pointing down, thumb pointing north but 30° below the horizontal (bend your thumb down without moving your fingers). Your palm faces west, so the force is east.
(d) Fingers pointing down, thumb pointing up ... ouch! The pain reminds you that if the velocity is parallel (or anti-parallel) to the magnetic field then the force is zero. A zero force has no direction, and that's the answer.
A duck flying horizontally due north at 15m/s passes over Atlanta, where the magnetic field of the Earth is 5.0×10-5T in a direction 60° below a horizontal line running north and south. The duck has a positive charge of 4.0×10-8C. What is the magnetic force acting on the duck?
As always, my advice is to begin with a picture.
A side view (looking east) should suffice.
Indicate the duck, moving north, and the direction of B (B points northward, at an angle of 60° from horizontal).
Now use F=qvBsinq, with q = 60° (your picture should show that the angle between v and B is 60°):
F = (4.0×10-8C)(15m/s)(5.0×10-5T)sin 60° = 2.6×10-11N.
Using the right and rule, fingers pointing north, 60° below horizontal, thumb pointing north, palms are facing to the west.
Since the charge on the duck is positive, the force points to the west.
A current is the collective motion of charges. Therefore, if a wire carrying a current is placed in a magnetic field, the individual charges will feel a force, and they will transmit this force to the wire. If the wire carries current I, and has length l in a magnetic field, B, and the current (the wire) makes an angle q with the magnetic field, the resulting force is:
Many practical devices involve a loop of wire in a magnetic field. A loop of wire will experience zero net force, but there will be a net torque. The operation of galvanometers, generators, and electric motors are based on the torque exerted on a current loop.
Consider a loop of wire in a magnetic field. For simplicity let the loop be rectangular, with two sides perpendicular to the field, and two sides parallel. The two perpendicular wires will carry equal currents but in opposite directions, therefore they will feel equal and oppositely directed forces. The wires parallel to the field feel no force. The net force is the sum of two equal and opposite forces and this is zero.
Recall from chapter 8 that the torque is defined as t = Fd, where F is the force and d is the lever arm (the distance perpendicular to the force, from the point of application of the force to the pivot axis). (The SI unit for torque is newton-meter, Nm.) Let's take the pivot axis to be through center of the parallel wires, halfway between the applied forces. Then we see that both applied forces tend to rotate the loop in the same direction, resulting in a net torque. The general expression we will use for the torque is:
A 2.00m long wire carrying a current of 2.00A forms a 1 turn loop in the shape of an equilateral triangle. If the loop is placed in a constant magnetic field of magnitude 0.500T, determine the maximum torque that acts on it.
The maximum torque is tmax = BIA, where I've set N=1 (1 turn loop) and sinq = 1. B and I are given. The trick here is to determine A from the information given. From what we are told, we must determine the area of an equilateral triangle whose perimeter is 2.00m. The length of the sides is equal to one third of the perimeter, or 0.667m. The area is ½base×height, with the base = 0.667m and the height = (0.667m)sin 60° = 0.577m, or A = ½(0.667m)(0.577m) = 0.192m². Therefore, tmax = BIA = (0.500T)(2.00A)(0.192m²) = 0.192 Nm.