Handout 8: Sources of magnetic field. Magnetic field of moving charge

Transcription

1 1 Handout 8: Sources of magnetic field Magnetic field of moving charge Moving charge creates magnetic field around it. In Fig. 1, charge q is moving at constant velocity v. The magnetic field at point P, distance r from the charge, is B = μ 0 qv r r 2, where μ 0 is a constant called permeability and its value in free space is equal to 10 7 TmA -1. The unit vector r is drawn from the charge to the point P. Note that at any point along the line of motion of the charge, v r = 0 and hence there is no magnetic field along this line. The magnetic field circulates around the line of the path of motion of the charge. Figure 1: Magnetic field due to moving charge Example 1 A proton of charge q = C is moving along a straight line with constant speed v = ms -1. Point P is at distance r = 1 m from the proton. The line joining P to the charge makes angle θ = 30 to the line of proton s motion. r P q θ v a) Find the electric field at point P. b) Find the magnetic field at point P. Example 2 Two protons move parallel to each other along x-axis in the opposite directions at the same speed v. Show that the ratio of magnetic force to electric force on each proton is equal to ε 0 μ 0 v 2.

2 2 Biot-Savart law The source of magnetic field is moving charge, i.e., electric current. Figure 2 shows a current element of length dl producing a magnetic field element db at point P. Inside dl, charge element dq is moving at drift speed v d. Therefore, db = μ 0 dq v d r r 2. Note that dq v d = Idt v d = Idl. The total magnetic field at point P is the integral over the whole length of the wire. We arrive at the Biot and Savart law: Figure 2: Magnetic field of current element B = μ 0 Idl r r 2. The direction of the magnetic field around a current-carrying conductor can be found by using right hand as shown in Fig. 3. The thumb points along the current and the four fingers curl in the direction of the magnetic field. Fig. 4 shows the directions of magnetic field produced by current in a circular loop. Figure 3: The magnetic field around a current-carrying conductor Infinitely long straight wire Let us demonstrate the use of Biot-Savart law by finding the magnetic field produced by an infinitely long straight wire carry a current. Figure 5 shows a current I flowing along an infinitely long wire. Point P is at distance R from the wire. The current element of length dx produces, at point P, the field element, db = μ 0I dx sin φ r 2 = μ 0 I dx cos θ r 2 Figure 4: The magnetic field produced by current in a circular loop in the direction out of the page. By writing cos θ = R x 2 + R 2 and r = x 2 + R 2, we have db = μ 0I Rdx x 2 + R 2 3/2. To obtain the magnetic field at point P, the above equation is integrated from x = to x = (the wire is infinitely long). Therefore, Figure 5: Current element dx producing field element db at point P

3 3 B = μ 0IR dx x 2 + R 2 3/2. By using simple integration technique, the integral is equal to 2 R 2. The magnetic field produced by an infinitely long current-carrying straight wire is given by B = μ 0I 2πR. Example 3 Let the current I flow around the loop of radius R. Use Biot-Savart law to show that the magnetic field at the center of the loop is given by B = μ 0I 2R. Force between parallel conductors Figure 6 shows two parallel conductors of equal length l separated by distance d (d l). The wire 1 carries the current I 1 and the 2 carries the current I 2 in the same direction. Consider the magnetic field B 2 at wire 1, produced by wire 2. We have B 2 = μ 0 I 2 2πd. Wire 1 is in the field B 2 and hence experiences magnetic force F 1 towards wire 2. The magnitude of the force is given by F 1 = B 2 I 1 l = μ 0 I 2 2πd I 1l = μ 0I 2 I 1 l 2πd. Figure 6: Two parallel conductors with current flowing in the same direction If we consider the force F 2 acting on wire 2 due to the magnetic field from wire 1, we find that F 2 = F 1 but in opposite directions. The wires attract. The same argument can be applied to case when the currents are in the opposite directions. The result is that the wires repel.

4 4 Ampere s law Ampere s law is formulated in terms of the line integral of magnetic field. Consider the line integral of B along the path in Fig. 7: B cos θ dl = B dl. The line integral is expressed as a dot product. Ampere s law state that the line integral of B around the a closed path is equal to the current enclosed by that path multiplied by μ 0 : Figure 7: Line integral of magnetic field along a chosen path B dl = μ 0 I enc. The symbol denotes integral around a closed path. The application of Ampere s law is particularly useful in system with certain symmetry. The use of Ampere s law follow the following steps Draw Amperian loop such that the magnetic field is aligned with the loop and the magnitude of the magnetic field is constant along the loop. Evaluate B dl = B dl. Find the enclosed current I enc. Find magnetic field B. Applications of Ampere s law Infinitely long conductor The magnetic field around the a long straight conductor is circular. Draw an Amperian loop to be a circle of radius r centered at the conductor as shown in Fig. 8. Vector B is parallel to dl and the magnitude of B is constant along the loop. Therefore, B dl = B dl = B dl = B 2πr. The enclosed current is the current I of the conductor. We have B 2πr = μ 0 I B r = μ 0 I 2πr. Figure 8: Amperian loop around a long straight wire The result is the same as that obtained from Biot-Savart law.

5 5 Solenoid The magnetic field produced by a solenoid is shown in Fig. 9. The field is approximately uniform inside the solenoid and zero outside. We use Ampere s law to find the field inside the solenoid by drawing Amperian loop abcd as illustrated in Fig. 10. The path bc and da are perpendicular to the field and the line integral is zero. The path cd is outside the solenoid where the field is zero. The only nonzero line integral is along ab which gives Figure 9: Magnetic field by solenoid B dl = B dl = B dl = BL. The enclosed current I enc = NI where I is the current in the coil and N is the number of turns over the length L. Therefore, from Ampere s law, BL = μ 0 NI B = μ 0NI L. The above expression is often written in terms n = N/L which is the number of turns per unit length. Figure 10: Amperian loop around the solenoid Toroid Toroid is a donut-shaped coil as shown in Fig. 11. The current I passes through the coil. The magnetic field in the toroid circulates and its magnitude is constant. Consequently, it is helpful to draw an Amperian loop as the circle of radius r. The line integral along this loop becomes B dl = B dl = B dl = B 2πr. The enclosed current I enc = NI where N is the number of turns. From Ampere s law, B 2πr = μ 0 NI B = μ 0NI 2πr. Figure 11: Toroid *Example 4 A cylinder of radius R carries current I uniformly distributed over the cross-section of the cylinder. Find magnetic field B(r) at distance r from the cylinder axis. Distinguish the case r > R and r < R.