Teaching Magnetism AP Summer Institute in Physics

Sample Problem n A compass points to the Earth’s North Magnetic Pole (which is near the North Pole). n Is the North Magnetic Pole the north pole of the Earth’s Magnetic Field?

Magnetic Force on a Charged Particle n magnitude: F = qvBsin  –q: charge in Coulombs –v: speed in meters/second –B: magnetic field in Tesla –  : angle between v and B n direction: Right Hand Rule n F B = q v x B (This is a “vector cross product”)

The Right Hand rule to Determine a Vector Cross Product 1. Align your hand along the first vector. 2. Orient your wrist so that you can “cross” your hand into the second vector. 3. Your thumb gives you the direction of the third vector (which is the result).

Sample Problem: Calculate the magnitude force exerted on a 3.0  C charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed a) North. b) South. c) East. d) West.

Sample Problem n Calculate the magnitude and direction of the magnetic force. v = 300,000 m/s B = 200 mT q = 3.0  C 34 o

Magnetic forces… n are always orthogonal (at right angles) to the plane established by the velocity and magnetic field vectors. n cause centripetal acceleration n cannot change the speed or kinetic energy of charged particles. n cannot do work on charged particles.

B V Sample Problem: Do a Newton’s 2 nd Law analysis of the motion of this positively charged particle in a magnetic field.

Sample Problem: Calculate the force and describe the path of this electron. E = 2000 N/C e-e- 300,000 m/s

Sample problem: Calculate the force and describe the path of this electron. e-e- 300,000 m/s B = 2000 mT

Sample Problem: It is found that protons traveling at 20,000 m/s pass undeflected through the velocity filter below. What is the magnitude and direction of the magnetic field between the plates? 400 V +e 20,000 m/s

Magnetic Force on Current- Carrying Wire n F = I L B sin  –I: current in Amps –L: length in meters –B: magnetic field in Tesla –  : angle between current and field n Derive this!

Sample Problem: What is the magnetic field strength if the current in the wire is 15 A and the force is downward and has a magnitude of 40 N/m? What is the direction of the current?

Magnetic Field produced by straight currents. I

Right Hand Rule for straight currents 1. Curve your fingers 2. Place your thumb (which is presumably pretty straight) in direction of current. 3. Curved fingers represent curve of magnetic field. 4. Field vector at any point is tangent to field line. i r 

Magnitude of Magnetic Field for Straight currents. n B =  o I / (2  r) –  o : 4   T m / A magnetic permeability of free space –I: current (A) –r: radial distance from center of wire (m)

Sample problem: what is the magnitude and direction of the force exerted on a 100 m long wire that passes through point P which bears a current of 50 amps in the same direction? I 1 = 13.0 A P 3.0 m I 2 = 50.0 A

Sample Problem: Where would the magnetic field be zero? 7.0 m I 1 = 10.0 A I 2 = 13.0 A

Right Hand Rule for magnetic fields around curved wires 1. Curve your fingers. 2. Place them along wire loop so that your fingers point in direction of current. 3. Your thumb gives the direction of the magnetic field in the center of the loop, where it is straight. 4. Field lines curve around and make complete loops. I B

Sample problem: What is the direction of the magnetic field inside the air-core solenoid shown? 100-  120 V

Magnetic Flux n The product of magnetic field and area. n Can be thought of as a total magnetic “effect” on a coil of wire of a given area. B A

Area Vector n Consists of the measured or calculated area (the vector’s magnitude) with a direction normal to the area’s surface.

Maximum Flux n Can you draw the loop so that the flux is maximum through the area? Make sure you draw the area vector! B A

Minimum Flux n Can you draw the loop so that the flux is minimum through the area? Make sure you draw the area vector! B A

Intermediate Flux n Can you draw the loop so that the flux is neither maximum nor minimum through the area? Make sure you draw the area vector! B A

Magnetic Flux  B = B A cos  –  B : magnetic flux in Webers (Tesla meters 2 ) –B: magnetic field in Tesla –A: area in meters 2. –  : the angle between the area and the magnetic field.  B = B  A

Induced Electric Potential n A system will respond so as to oppose changes in magnetic flux. n A change in magnetic flux will be partially offset by an induced magnetic field whenever possible. n Changing the magnetic flux through a wire loop causes current to flow in the loop. n This is because changing magnetic flux induces an electric potential.

Faraday’s Law of Induction –  : induced potential (V) –N: # loops –  B : magnetic flux (Webers, Wb) –t: time (s)  To generate voltage, change B, change A, change 

Lenz’s Law n The current will flow in a direction so as to oppose the change in magnetic flux. n Use in combination with hand rule to predict current direction.

Sample Problem: The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop?

Motional emf  = BLv –B: magnetic field (T) –L: length of bar moving through field –v: speed of bar moving through field. n Bar must be “cutting through” field lines. It cannot be moving parallel to the field. n This formula is easily derivable from Faraday’s Law of Induction

Sample Problem: How much current flows through the resistor, and in what direction? How much power is dissipated by the resistor?       B = 0.15 T v = 2 m/s 50 cm 3 