1. Review: Magnetic Force on Current-Carrying Wire
F = I ℓ B
I: current in Amps
ℓ : length in meters
B: external magnetic field in Tesla
direction: Right Hand Rule II

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

3. Magnetic Field produced by straight currents.

4. F = I ℓ B 40 = (15) B B = 2.7 T right F / ℓ = IB
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?
F = I ℓ B
40 = (15) B
B = 2.7 T
right
F / ℓ = IB

5. 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?
I1 = 13.0 A P
3.0 m
I2 = 50.0 A
µo I
4π x 10-7(13)
B =
B =
B = 8.6 x 10-7 T out of the page
2r
2(3)
F = I ℓ B
F = 50 (100) (8.6 x 10-7)
F = 4.3 x 10-3 N down
Field due to bottom wire at the top wire

6. Review: Magnetic Force on a Charged Particle
magnitude: F = qvB
q: charge in Coulombs
v: speed in meters/second
B: external magnetic field in Tesla
direction: Right Hand Rule II

7. North. South. East. West. F = 180 N into the page
Sample problem: Calculate the magnitude force exerted on a 3.0 mC charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed
North.
South.
East.
West.
Zero
Zero
F = qvB
F = 180 N into the page
F = (3x10-3)(300,000)(200 x 10-3)
F = qvB
F = 180 N out of the page
F = (3x10-3)(300,000)(200 x 10-3)

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

9. Sample problem: Calculate the force and describe the path of this electron.
B = 200 mT
x x x x x x x x x
v = 300,000 m/s e-
F = qvB
F = (1.6x10-19)(300,000)(200 x 10-3)
It curves like this with radius :
F = 9.6 x N initially down (always towards the center)
r = mv/qB
8.5 x 10-6 m
r = 9.1 x 10-31(300,000)/(1.6 x 10-19)(200 x 10-3)

10. B = 0.653 T +e into the page 2.1 x 10-15 = (1.6 x 10-19)(20,000) B
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
B = T
3 cm
into the page
2.1 x = (1.6 x 10-19)(20,000) B
F = qvB
V = Ed
F = Eq
400 = E(0.03)
F = 13,333(1.6 x 10-19)
13,333 N/C
F = 2.1 x N

11. Undeviated path
As we just discussed, for a moving electric charge to pass through a magnetic field undeflected, an additional force, equal and ______________ to the magnetic force must be acting on the charge.
opposite
If this additional force is caused by an electric field:
F = Eq
Eq = qvB or E = vB

12. Equations
E = vB*
*undeviated path