Inducing an Emf (Voltage Difference) When a collection of positive and negative charges move at some speed v through a perpendicular magnetic field, of course, the negative charges get pushed one way; the positive charges get pushed the other way. This is exactly what happens, too, when those charges are in a conductor—say, a bar of length L—that is moving at that speed v through that B-field: The magnetic force on an electron will push it to one end of the bar until there is such a negative charge build-up there that the magnetic force is equalized by the electric (repulsion) force of all the other extra electrons already gathered there. And of course, the converse is true at the other end of the bar—net positive charge increases. 5/24/17 OSU PH 213, Before Class #23

When this force balance is achieved, we have a lot of positive charge on one end of the bar and a lot of negative charge on the other end. In other words, it looks a lot like a battery. And we can calculate the voltage difference, V (also called the emf) between the two ends of the bar, simply by equating the magnetic and electric forces: qvB = qE Now recall that one way to express an electric field is E = –V/s. If we take the magnitude information there (E = V/s), substitute it into the equation above and solve for V, we get V = vBs. In other words: V = vBL 5/24/17 OSU PH 213, Before Class #23

What if the charges being separated by the magnetic force are not stopped at the ends of the bar? What if that bar is attached to a circuit? Then, just as with a battery, the charges flow around the circuit—and all the usual circuit calculations apply.… 5/24/17 OSU PH 213, Before Class #23

Notice: When there is a current flowing through the bar, there is a force on that bar, which (since the bar is perpendicular to the B-field) is given by F = ILB. And the direction of that force is opposite the motion of the bar. So, to keep the bar moving at the steady speed, v, we must push it along with an equal and opposite force. This force, acting in the same direction as the velocity of the bar, does work—at a rate equal to the power dissipated in the circuit: ILB·v = V·I 5/24/17 OSU PH 213, Before Class #23

Now put it all together.… What’s the point here? Nobody goes around with a conductive bar trying to power light bulbs by running through magnetic fields. Never mind the math of this particular situation. What’s the underlying principle? To get an idea, look again at the form of the voltage difference that arises due to the motion: vLB = V Recall that v = x/t, and notice that the area, A, of the rectangular loop is given by L·x (and L isn’t changing here). Notice, too, that the voltage difference appeared even when current could not flow. As a matter of fact, it turns out that no charge even needs to be present for that voltage difference to arise. Now put it all together.… 5/24/17 OSU PH 213, Before Class #23

Changing Magnetic Flux: Faraday’s Law Mathematically, we can express the motion-induced voltage in any closed path as the rate of change (over time) of the area of that loop times the perpendicular B-field passing through that area. That is (in non-differential form): |V| = [(B·cos)A]/t M is called the magnetic flux of the loop: M = B•A = (B·cos)A where  is the angle between B and the normal to the loop of area A. The SI unit of magnetic flux is the Weber (Wb). 1 Wb = 1 T·m2 Written as an instantaneous value, we then have: |V(t)| = d/dt[M] 5/24/17 OSU PH 213, Before Class #23

M = (B·cos)A Suppose that at a certain location the earth’s natural magnetic field is 5 x 10-5 T, directed into the ground at an angle of 60° below the positive horizontal. What is the magnetic flux through a circular ring of copper (2 m in diameter) which is lying flat on level ground? A. 6.28 x 10-5 Wb B. 1.36 x 10-4 Wb C. 2.51 x 10-4 Wb D. 5.44 x 10-4 Wb E. None of the above. 5/24/17 OSU PH 213, Before Class #23

M = (B·cos)A Suppose that at a certain location the earth’s natural magnetic field is 5 x 10-5 T, directed into the ground at an angle of 60° below the positive horizontal. What is the magnetic flux through a circular ring of copper (2 m in diameter) which is lying flat on level ground? A. 6.28 x 10-5 Wb B. 1.36 x 10-4 Wb C. 2.51 x 10-4 Wb D. 5.44 x 10-4 Wb E. None of the above. 5/24/17 OSU PH 213, Before Class #23

What Can Change the Magnetic Flux? The changing nature of a loop (area or angle) in a magnetic field is not the only way to change the flux  = (B·cos)A. There’s a third variable: Even if the loop remains motionless and unchanging, the flux can change if the magnitude or direction of the magnetic field, B, is changing. Faraday’s Law states that any of these will cause an induced voltage in a closed loop whose area is intersected by a magnetic field, B: ･ Change the area of the loop; ･ Change the orientation of the loop; ･ Change the magnitude of B; ･ Change the direction of B. 5/24/17 OSU PH 213, Before Class #23