Applied Physics Lecture 14 Electricity and Magnetism Magnetism Ampere’s law Applications of magnetic forces Chapter 19 4/26/2017

19.7 Motion of Charged Particle in magnetic field Bin Consider positively charge particle moving in a uniform magnetic field. Suppose the initial velocity of the particle is perpendicular to the direction of the field. Then a magnetic force will be exerted on the particle… ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ ´ F v q r Where is it directed? … and make it follow a circular path. Remember that 4/26/2017

The magnetic force produces a centripetal acceleration.   The particle travels on a circular trajectory with a radius: 4/26/2017

Example 1 : Proton moving in uniform magnetic field A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. Given: r = 0.14 m B = 0.35 T m = 1.67x10-27 kg q = 1.6 x 10-19 C Recall that the proton’s radius would be Thus Find: v = ? 4/26/2017

19.8 Magnetic Field of a long straight wire Danish scientist Hans Oersted (1777-1851) discovered (somewhat by accident) that an electric current in a wire deflects a nearby compass needle. In 1820, he performed a simple experiment with many compasses that clearly showed the presence of a magnetic field around a wire carrying a current. I=0 I 4/26/2017

Magnetic Field due to Currents The passage of a steady current in a wire produces a magnetic field around the wire. Field form concentric lines around the wire Direction of the field given by the right hand rule. If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of the field (second right-hand rule). Magnitude of the field I 4/26/2017

Magnitude of the field I r B mo called the permeability of free space 4/26/2017

Ampere’s Law Consider a circular path surrounding a current, divided in segments Dl, Ampere showed that the sum of the products of the field by the length of the segment is equal to mo times the current. Andre-Marie Ampere I r Dl B 4/26/2017

Consider a case where B is constant and uniform: Then one finds: 4/26/2017

l d 1 2 F1 B2 I1 I2 Force per unit length 4/26/2017

Definition of the SI unit Ampere Used to define the SI unit of current called Ampere. If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10-7 N/m, then the current is defined to be 1 A. 4/26/2017

Example 1: Levitating a wire Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current). l 1 I1 2 d I2 4/26/2017

mg/l = 1.0x10-4 N/m F1 1 I1 B2 mg/l 2 d I2 l Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current). 1 I1 B2 mg/l 2 d I2 l Weight of wire per unit length: mg/l = 1.0x10-4 N/m Wire separation: d=0.1 m I1 = I2 4/26/2017

19.10 Magnetic Field of a current loop Magnetic field produced by a wire can be enhanced by having the wire in a loop. Dx1 I B Dx2 4/26/2017

19.11 Magnetic Field of a solenoid Solenoid magnet consists of a wire coil with multiple loops. It is often called an electromagnet. 4/26/2017

Solenoid Magnet ali.riaz@uettaxila.edu.pk Field lines inside a solenoid magnet are parallel, uniformly spaced and close together. The field inside is uniform and strong. The field outside is non uniform and much weaker. One end of the solenoid acts as a north pole, the other as a south pole. For a long and tightly looped solenoid, the field inside has a value: 4/26/2017

Solenoid Magnet n = N/l : number of (loop) turns per unit length. I : current in the solenoid. 4/26/2017

Example: Magnetic Field inside a Solenoid. Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it carries a current of 0.500 A. N = 100 l = 0.100 m I = 0.500 A 4/26/2017

Comparison: Electric Field vs. Magnetic Field Electric Magnetic Source Charges Moving Charges Acts on Charges Moving Charges Force F = Eq F = q v B sin(q) Direction Parallel E Perpendicular to v,B Field Lines Opposites Charges Attract Currents Repel 4/26/2017