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.
A galvanometer makes use of the torque on a current loop placed in the field of a permanent magnet and can be used to measure current or voltage. A spring is used to make the deflection proportional to the current in the loop, and a needle is attached to indicate how far the loop rotates. A typical galvanometer has an internal resistance, r0, of about 50W, and deflects to its maximum for a current of a few milliamps (mA). This is not directly suitable as a current or voltage measuring device, as an ideal current meter (ammeter) has zero internal resistance, and an ideal voltmeter has infinite internal resistance. (Real ammeters have very small, but not zero, internal resistance, and real voltmeters have very large, but not infinite, internal resistance.) The following example demonstrates how a galvanometer is made part of an ammeter and voltmeter.
A 50.0W, 10.0mA galvanometer is to be converted to an ammeter that reads 3.00A at full-scale deflection. (a) What value of Rp should be placed in parallel with the coil? (b) What value of Rs should be placed in series with the coil to convert it to a voltmeter that reads 10.0V at full-scale deflection?
(a) To convert a galvanometer to an ammeter, we place a resistor in parallel with the coil.
Then part of the current will pass through the coil and the remainder through the parallel resistor, Rp.
The specification of the galvanometer tells us that the internal resistance r0 = 40.0W, and that 10.0mA causes the needle to deflect full-scale.
Therefore, when 3.00A of current pass through the coil and resistor in parallel, we want 10.0mA to pass through the coil, and the remaining 2.99A to pass through Rp.
Because the coil and resistor are in parallel, the same voltage appears across both.
When 10.0mA passes through the 50.0W coil, the voltage across it is V = IR = (10.0mA)(50.0W) = 500mV = 0.500V.
With this same voltage applied across it, Rp is to carry a current of 2.99A, therefore
Rp = V/I = (0.500V)/(2.99A) = 0.167W.
Note that this ammeter does indeed have a small effective resistance of about 0.165W.
(b) To convert a galvanometer to a voltmeter, we place a resistor in series with the coil.
The same current passes through the series resistor, Rs, and coil, the added resistance limits the current through the coil, and increases the effective resistance of the device.
If the needle is to deflect full-scale at a voltage of 10.0V, then the current through resistor and coil must be 10.0mA.
Thus the total resistance of the device is Rtot = Rs + Rcoil = V/I = (10.0V)/(10.0mA) = 1.00kW.
Since the coil has a resistance of 50.0W, we find
Rs = 1000W - 50.0W = 950W.
When a charged particle moves perpendicularly to a magnetic field, it feels a force F = qvB directed perpendicular to its motion and the magnetic field. This causes the charge to change direction, but doesn't affect is speed, v. The charge will follow a circular path. The force F is directed towards the center of the circle, and equals the centripetal force needed to keep the charge moving on the circle:
Consider the mass spectrometer shown schematically in Figure P19.32. The electric field between the plates of the velocity selector is 950V/m, and the magnetic fields in both the velocity selector and the devlection chamber have magnitudes of 0.930T. Calculate the radius of the path in the system for a singly charged ion with mass m=2.18×10-26kg.
The velocity selector is a device which uses both electric and magnetic fields to select particles moving at particular velocities.
Basically, particles can pass through the selector only if they have a velocity such that the magnetic force balances the electric force:
qvB = qE : or v = E/B.
So, in this device,
v = E/B = (950V/m)/(0.930T) = 1020m/s.
If the particle is singly charged, then q = 1.6×10-19C, and
r = mv/qB = (2.18×10-26kg)(1020m/s)/(1.6×10-19C)(0.930T) = 1.5×10-4m = 0.15mm.