1. 7.2 Magnetic Field Strength p. 274 Calculating Magnetic Field Strength A moving charged particle that enters a magnetic field at any direction other than parallel will experience a force, F B, that depends on the charge, Q, and speed, v, of the particle, the strength of the magnetic field, B. FBFB QvQv = constant = B F B = QvB Unit for magnetic Field strength is the Tesla (T): 1 N Cm/s 1 N Am = = 1 T If the charged particle’s motion is not at right angles to the magnetic field then the equation above must be modified: F B = QvB sin Ɵ

2. 7.2 Magnetic Field Strength The Magnetic Field Strength of a Solenoid (part 1) p. 276 - 277 Andrè Ampere investigated the magnetic fields associated with solenoids. He discovered how the magnetic field forms around these solenoids The strength, B, of the magnetic field inside a solenoid depends on: 1)The number of a turns, N, per unit length, L, of solenoid 2)The current in the solenoid I. NI B α L

3. 7.2 Magnetic Field Strength The Magnetic Field Strength of a Solenoid (part 2) p. 277 constant xB = NI L The constant of proportionality is given the name of permeability of free space. permeability of free space:µ o = 4π x 10 -7 Tm/A B = µ o NI L B = µ o nI L N n = Where: (n = number of turns per unit length) Inserting a constant changes the proportionality into an equation.

4. 7.2 Magnetic Field Strength Magnetic Force on a Current in a Wire (part 1) p. 278 A current carrying wire has a magnetic field. When this current carrying wire is placed in another external magnetic field, the wire will experience a force due to the interactions of the two magnetic fields. The right hand rule is also applied here to determine the force that exists on the current carrying wire. Point your right thumb in the direction of conventional current. Point your fingers straight out in the direction of the magnetic field. The palm of your right hand will now be pushing in the direction of the force on the wire segment carrying current in the magnetic field.

5. 7.2 Magnetic Field Strength Magnetic Force on a Current in a Wire (part 2) p. 279 Remember: F B = QvB And: v = d/t = l /t F B = BQ( l /t) And: I = Q/t So: Q = It F B = BIt( l /t) F B = BI l Therefore: If the wire is not perpendicular to the magnetic field then the component of the magnetic field perpendicular to the wire segment must be used. F B = BI l sin Ɵ

6. 7.2 Magnetic Field Strength Key Questions In this section, you should understand how to solve the following key questions. Page 276 – Quick Check #3 Page 277 – Quick Check #2 & 3 Page 282 – 283 – Review 7.2 #1,3,7,8, & 9