Chapter 29 Magnetic Fields Due to Currents In this chapter we will explore the relationship between an electric current and the magnetic field it generates in the space around it. We will follow a two-pronged approach, depending on the symmetry of the problem. For problems with low symmetry we will use the law of Biot-Savart in combination with the principle of superposition. For problems with high symmetry we will introduce Ampere’s law. Both approaches will be used to explore the magnetic field generated by currents in a variety of geometries (straight wire, wire loop, solenoid coil, toroid coil).We will also determine the force between two parallel, current-carrying conductors. We will then use this force to define the SI unit for electric current (the ampere). (29-1) 0i ds r dB 4 r 3 A The Law of Biot - Savart This law gives the magnetic field dB generated by a wire segment of length ds that carries a current i. Consider the geometry shown in the figure. Associated with the element ds we define an associated vector ds that has magnitude equal to the length ds. The direction of ds is the same as that of the current that flows through segment ds. The magnetic field dB generated at point P by the element dS located at point A i ds r is given by the equation dB 0 . Here r is the vector that connects 3 4 r point A (location of element ds) with point P at which we want to determine dB. The constant 0 4 10 7 T m/A 1.26 10 6 T m/A and is known as the "permeability constant." The magnitude of dB is dB Here is the angle between ds and r . 0i ds sin . 2 4 r (29-2) Magnetic Field Generated by a Long Straight Wire The magnitude of the magnetic field generated by the wire at point P located at a distance R from the wire is given by the equation i B 0 2 R 0i B . 2 R The magnetic field lines form circles that have their centers at the wire. The magnetic field vector B is tangent to the magnetic field lines. The sense for B is given by the right hand rule. We point the thumb of the right hand in the direction of the current. The direction along which the fingers of the right hand curl around the wire gives the direction of B. (29-3) B 0i 2 R Consider the wire element of length ds shown in the figure. The element generates at point P a magnetic field of i ds sin magnitude dB 0 . Vector dB is pointing 2 4 r into the page. The magnetic field generated by the whole wire is found by integration: 0i ds sin B dB 2 dB 2 2 r 0 0 r s2 R2 sin sin R / r R / s 2 R 2 0i 0i 0i Rds s B 2 3/2 2 2 2 2 0 s R 2 R s R 0 2 R dx x a 2 2 3/2 x a2 x2 a2 (29-4) B . Magnetic Field Generated by dB . a Circular Wire Arc of Radius R at Its Center C 0i B 4 R A wire section of length ds generates at the center C a magnetic field dB. 0i ds sin 90 0i ds The magnitude dB . The length ds Rd 2 2 4 R 4 R i dB 0 d . Vector dB points out of the page. 4 R i i The net magnetic field B dB 0 d = 0 . 4 R 4 R 0 Note : The angle must be expressed in radians. For a circular wire, 2 . In this case we get: Bcirc 0i 2R . (29-5) Ampere's Law The law of Biot-Savart combined with the principle of superposition can be used to B ds i 0 enc determine B if we know the distribution of currents. In situations that have high symmetry we can use Ampere's law instead, because it is simpler to apply. Ampere's law can be derived from the law of Biot-Savart, with which it is mathematically equivalent. Ampere's law is more suitable for advanced formulations of electromagnetism. It can be expressed as follows: The line integral B ds of the magnetic field B along any closed path is equal to the total current enclosed inside the path multiplied by 0 . The closed path used is known as an "Amperian loop." In its present form Ampere's law is not complete. A missing term was added by Clark Maxwell. The complete form of Ampere's law will be discussed in Chapter 32. (29-7) B ds i 0 enc Implementation of Ampere's Law : B ds . 1. Determination of The closed path is divided into n elements s1 , s2 , ..., sn . We then form the sum: n n B s B s cos . i 1 i i i i 1 i i Here Bi is the magnetic field in the ith element. n B ds lim B s i 1 i i as n 2. Calculation of ienc . We curl the fingers of the right hand in the direction in which the Amperian loop was traversed. We note the direction of the thumb. All currents inside the loop parallel to the thumb are counted as positive. All currents inside the loop antiparallel to the thumb are counted as negative. All currents outside the loop are not counted. (29-8) In this example : ienc i1 i2 . Magnetic Field Outside a Long Straight Wire We already have seen that the magnetic field lines of the magnetic field generated by a long straight wire that carries a current i have the form of circles, which are concentric with the wire. We choose an Amperian loop that reflects the cylindrical symmetry of the problem. The loop is also a circle of radius r that has its center on the wire. The magnetic field is tangent to the loop and has a constant magnitude B : B ds Bds cos 0 B ds 2 rB i 0 enc B 0i 0i 2 r Note : Ampere's law holds true for any closed path. We choose to use the path that makes the calculation of B as easy as possible. (29-9) Magnetic Field Inside a Long Straight Wire We assume that the distribution of the current within the cross-section of the wire is uniform. The wire carries a current i and has radius R. We choose an Amperian loop that is a circle of radius r (r R) with its center on the wire. The magnetic field is tangent to the loop and has a constant magnitude B : B ds Bds cos 0 B ds 2 rB i 0 enc ienc B r2 i 2 rB 0i 2 B 0 2 r R 2 R 0i 2 R O r2 r2 i i 2 2 R R R r (29-10) The Solenoid The solenoid is a long, tightly wound helical wire coil in which the coil length is much larger than the coil diameter. Viewing the solenoid as a collection of single circular loops, one can see that the magnetic field inside is approximately uniform. The magnetic field inside the solenoid is parallel to the solenoid axis. The sense of B can be determined using the right-hand rule. We curl the fingers of the right hand along the direction of the current in the coil windings. The thumb of the right hand points along B. The magnetic field outside the solenoid is much weaker and can be (29-11) taken to be approximately zero. We will use Ampere's law to determine the magnetic field inside a solenoid. We assume that the magnetic field is uniform inside the solenoid and zero outside. We assume that the solenoid has n turns per unit length. B 0 ni We will use the Amperian loop abcd . It is a rectangle with its long side parallel to the solenoid axis. One long side (ab) is inside the solenoid, while the other (cd ) is outside: b c d a a b c d B ds B ds B ds B ds B ds b b b a a a B ds Bds cos 0 B ds Bh B ds Bh B ds i 0 enc c d a b c d B ds B ds B ds 0 The enclosed current ienc nhi. Bh 0 nhi B 0 ni (29-12) B o Ni 2 r Magnetic Field of a Toroid A toroid has the shape of a doughnut (see figure). We assume that the toroid carries a current i and that it has N windings. The magnetic field lines inside the toroid form circles that are concentric with the toroid center. The magnetic field vector is tangent to these lines. The sense of B can be found using the right-hand rule. We curl the fingers of the right hand along the direction of the current in the coil windings. The thumb of the right hand points along B. The magnetic field outside the solenoid is approximately zero. We use an Amperian loop that is a circle of radius r (orange circle in the figure): B ds Bds cos 0 B ds 2 rB. Thus: 2 rB 0 Ni B The enclosed current ienc Ni. 0 Ni . 2 r Note : The magnetic field inside a toroid is not uniform. (29-13) The Magnetic Field of a Magnetic Dipole. Consider the magnetic field generated by a wire coil of radius R that carries a current i. The magnetic field at a point P on the z -axis is given by B B( z ) 0 3 2 z 0iR 2 2R z 2 2 3/2 . Here z is the distance between P and the coil center. For points far from the loop (z B R) we can use the approximation: B 0i R 2 0iR 2 2z3 0iA 0 . Here is the magnetic 3 3 2 z 2 z 2 pz 3 dipole moment of the loop. In vector form: B( z ) 0 . 3 2 z The loop generates a magnetic field that has the same form as the field generated by a bar magnet. (29-14) Example 1: Example 2: Example 3: Example 4: Example 5: Example 6: Example 7: Example 8: Example 9: Example 10: Example 11: Example 12: Example 13: Example 14: Example 15: Example 16: Example 17: Example 18: Example 19: Example 20: Example 21: Example 23: Example 24: