University:
California State University, Northridge
Course:
PHYS 220B | Electricity and Magnetism
Academic year:

2013
Views:

478

Pages:

20
Author:

tusjm5f

Magnetism Outline 22-1 The Magnetic Field 22-2 The Magnetic Force on Moving Charges 22-3 The Motion of Charges Particles in Magnetic Field 22-4 The Magnetic Force Exert on a Current-Carrying Wire 22-5 Loops of Current and Magnetic Torque 22-6 Electric Current, Magnetic Fields, and Ampère’s Law 22-7 Electric Loops and Solenoid 22-5 Loops of Current and Magnetic Torque Rectangular Current Loops A rectangle loop has height h and width w, and carriesa current I, as shown. The loop is placed in a uniform magnetic field B . The magnetic force on each vertical side (h height) is F  IhB The loop is intend to rotate around its center. The torque  is w 2 w 2   ( IhB)( )  ( IhB)( )  IB(hw) Since area A= hw, The torque  is   IBA Torque applied on a rectangular Loop of Area A   IBA sin  (22  5) SI unit: N.m Note: the angle  is the angle between the loop plane and the magnetic force on each side of the loop; This equation is suitable for any shape loop. Torque applied on a rectangular Loop of Area A and N turns of loops   NIBA sin  SI unit: N.m (22  6) Example 22-5 Torque on a Coil A rectangular coil with 200 turns is 5.0 cm high and 4.0 cm wide. When the coil is placed in a magnetic field of 0.35 T, its maximum torque is 0.22 N.m. What is the current in the coil? Example 22-5 Torque on a Coil Solution: 1. For maximum torque, one has   NIBA 2. Solving for I, I  mas NAB 0.22 N .m   1.6 A (200)(0.050m)(0.040m)(0.35T ) 22-6 Electric Current, Magnetic Fields, and Ampère’s Law A Long, Straight Wire a) Iron powder re-arranged when I go through the wire. b) Compass point in same direction around a circle. Figure 22-19 The Magnetic Field of a Current-Carrying Wire Magnetic Field Right-Hand Rule The direction of magnet field induced by a current-carrying wire can be found by: point the thumb of your right hand along the wire in the current direction I. Your fingers then will curl along the direction of the magnetic field. Figure 22-20 The Magnetic-Field Right-Hand Rule The direction of the magnetic field can be determined by using Magnetic Field Right-Hand Rule Ampère’s Law Ampère’s Law relates to the magnetic field along a closed path to the electric current. For a straight line segment L, it can be expressed as (in the figure) Ampère’s Law  B L   I || 0 enclosed Where, 0 = 4 x 10-7 T.m /A, is a constant called Permeability of Free Space. For a circular (see Figure), the circumference is 2r. we have  B L B L B(2r ) || Therefore, we have B(2r )   0 I Figure 22-22 Applying Ampère’s Law Magnetic Field for a Long, Straight Wire 0 I B 2 r (22  9) SI unit: Tesla, T Where, 0 = 4 x 10-7 T.m /A, is a constant called Permeability of Free Space. Exercise 22-2 Find the magnitude of magnetic field 1 m away from a long straight wire carrying a current of 1 A. Solution 0 I (4 107 T  m / A)(1A) B   2 107 T 2 r 2 (1m) Example 22-6 An Attractive Wire A 52 uC charged particle parallel to a long wire with a speed of 720 m/s. The separation between the particle and the wire is 13 cm, and the magnitude of a force exerted on the particle is 1.4x10-7 N. Find (a) the magnitude of the magnetic field B at the location of the particle, and (b) the current I in the wire. Picture the problem Solution Part (a) Since F=qvB F 1.4 10 7 N B  qv (5.2 10 5 C )(720m / s )  3.7 10 6 T Part (b) Since 0 I B 2r Solving for I, we have I 2 rB 0 2 (0.13m)(3.7  106 T )   2.4 A 7 4 10 T  m / A Active Example 22-2 Find the magnetic Field Two wires separated by a distance of 22 cm carry currents in the same direction. The current in one wire is 1.5 A, and the current in the other is 4.5 A. Find the magnitude of the magnetic field halfway between the wires. Active Example 22-2 Find the Magnetic Field Solution: 1) Find the magnitude of the magnetic field of wire 1 B1 = 2.7x10-6 T, into page 2) Find the magnitude of the magnetic field of wire 2 B2 = 8.2x10-6 T, out of page 3) The net magnetic field B = B2 - B1 = 5.5x10-6 T Active Example 22-2 Find the Magnetic Field Forces between Current-Carrying Wires How large the force is? Figure 22-24 The Direction of the Magnetic Force Between Current-Carrying Wires Summary 1. Torque Exert on a rectangular Loop of Area A and N turns   NIBA sin  (22  6) 2. Magnetic Field for a Long, Straight Wire 0 I B 2 r (22  9)

Magnetism

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