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Chapter 19 Magnetic Force on Wires Solenoids Mass Spectrometers 0I B 2r The magnetic field around a very long, current carrying wire is given by: Where the constant of proportionality is known as the permeability of free space and is found to be 0 4 10 Tm / A 7 The magnetic field is a vector quantity, so it has a direction!!!! Wrap your hand around the wire with your thumb pointed in the direction of the current. I Your fingers curl around the wire in the direction the magnetic field points. If we have two very long, current carrying wires parallel to one another, what force do they feel? I1 I2 d F B IL sin = 90o Fmax B IL If the wires are parallel, the magnetic field from one wire will be perpendicular to the direction of the current in the other wire. I1 I2 1 d 2 F B1@2 I 2 L2 0I1I 2 F L2 2d 0 I1 B1@2 2d F 0I1I 2 L2 2d I1 I2 1 r 2 In what direction does the force act??? Using the right hand rule, we determine that the magnetic field from wire 1 at the location of wire 2…. Is into the page! Now, using the OTHER right hand rule (the one for magnetic force on a current carrying wire in a magnetic field), we find that the direction of the force on wire 2 ... Is UP toward wire 1! I1 I2 1 r 2 Parallel wires carrying current in the SAME direction are attracted to one another. You might be able to guess… Parallel wires carrying current in the OPPOSITE direction are repelled from one another. But what if the wire isn’t a very long, straight wire? What if it’s a loop of wire instead? We won’t go into the details of the magnetic field created by a single loop, but this is what it looks like --- a lot like the field of a bar magnet! However, we will look at the magnetic field created by a whole bunch of such loops combined…AKA Solenoids produce fairly uniform magnetic fields inside their boundaries, and generate negligible fields outside their boundaries. Binside 0nI Where n is the number of turns per unit length of the solenoid. The magnetic field is directed parallel to the axis of the solenoid. The right hand rule for currents will help you determine toward which end the field points. A solenoid has length of 25 cm, radius of 1 cm, 400 turns, and carries a current of 3 A. What is the value of the magnetic field inside the solenoid? Binside 0nI N 400 1 n 1600m L .25 7 3 Binside (4 10 )(1600)(3) 6 10 T We saw that charged particles in a uniform magnetic field will move in a circle. x x x x x x x x x x x x x r x x x x x x x We can take advantage of our understanding of the magnetic force to create a device which can tell us the elemental composition of unknown particles. x x x x x x x x x x x x x r x x x x x x x mv r qB We saw that the radius of the circle is determined by the mass, speed, and charge of the particles, and the strength of the magnetic field. If we know the velocity at which a particle enters a magnetic field, the strength of that field, and the particle’s charge, we can measure the radius of curvature of the particle. x x x x x x x x x x x x x x x x x x x x d rq B m v We can measure d using a photographic plate (for example) to record the location of the end of the particle’s path. d=2r If two particles with the same charge enter the magnetic field with the same velocity, then the difference in the radii of their paths must be due to a difference in their masses. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x d1 d2 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x mv r qB d1 d2 d 2 d1 v ( m2 m1 ) 2 2 qB 2v d 2 d1 (m2 m1 ) qB Mass Spectrometer Two singly ionized particles enter a mass spectrometer at a speed of 3 X 106 m/s. The strength of the magnetic field is 0.625 T. If one of the particles is H and the other particle hits the photographic plate 110 cm further away than the H atom, what chemical element is the second particle? Let’s first determine the distance from the entrance point that the H atom hits the photographic plate. 27 mv (167 . 10 kg)(3 10 m / s) r 5 cm 19 qB (16 . 10 C)(0.625T) 6 Now, let’s figure out the radius of curvature of the mystery element... We were told that the 2nd element collides with the photographic plate 110 cm further away from the entrance point than the H atom. x x x x x x x x x x x x x x x x x x d2 - d1= 110 cm x x x x x x x x x x x x x x x x x d2 = 110 + 2r1 x d2= 120 cm d1 d2 r2= 60 cm Finally, let’s determine the mass of our mystery element... 19 rqB (0.6 m)(16 . 10 C)(0.625T) 26 m 2 10 kg 6 v 3 10 m / s PHYSICS RULES! Televisions rely upon electromagnetic fields to produce the images we see. An electron “gun” fires electrons at the television screen. When they collide with the material on the back side of the screen, colored light is emitted, producing one pixel in the image we see on the screen. television screen electromagnet Changing the strength of the magnetic field changes the degree to which the electron beam is deflected. We now make explicit a fact we’ve assumed implicitly in dealing with magnetic fields thus far... The superposition principle applies! x 1 Btot = B1 + B2 x 2 Induced Voltages and Inductance We saw in Chapter 19 that moving charges (currents) create magnetic fields. Nature often reveals a great deal of symmetry. So, scientists wondered, “Can magnetic fields can create currents?” You might guess that the answer is, “Yes.” In this chapter, we’ll explore this physical process known as induction. Shortly after Oersted’s discovery that currents create magnetic fields, Michael Faraday (England) and Joseph Henry (US) conducted experiments to determine if magnetic fields created electrical currents... Soft magnetic material + A _ primary coil Primary Circuit secondary coil Secondary Circuit A + _ Faraday’s Observations: With the switch open, no reading on the ammeter. When the switch is closed, the ammeter needle deflects momentarily toward the right, then returns to 0. A + _ Faraday’s Observations: With a constant current flowing through the primary circuit, the ammeter shows no reading. When the switch is opened, the ammeter needle deflects to the left momentarily, then returns to 0. A + _ Faraday’s Conclusions: Steady currents in the primary produce constant magnetic fields, which result in no current in the secondary circuit. Changing currents in the primary circuit result in changing magnetic fields, which result in induced currents in the secondary circuit. A If we move a bar magnet toward a loop of wire, the ammeter deflects to the left. When we stop moving the magnet, the ammeter reads 0. When we pull the magnet away from the loop, the ammeter deflects to the right. A As the magnet moves toward the loop, the magnetic field near the loop increases. As the magnet moves away from the loop, the magnetic field near the loop decreases. Such results are consistent with Faraday’s experiment: Changing magnetic fields lead to induced currents. Magnetic Flux x x x x x x x x x x x x x x x x x x x x side view How much magnetic field is passing through the loop? One way to think about this is to simply count the number of magnetic field lines passing through a loop. x x x x x x x x x x x x x x x x x x x x 30o 60o top view side view Magnetic flux is the amount of magnetic field perpendicular to the loop. If you separate the magnetic field into components parallel and perpendicular to the loop, you see that... 30o 60o =| B |A =| B|A cos top view Btot B B|| This is also the direction of the normal to the loop! = B A = BA cos = B A cos Wb 2 T m 2 m Wb m 2 So, the Weber is the unit of magnetic flux.