Physics 4B Lecture Notes Chapter 30 - Magnetic Fields Due to Currents Problem Set #9 - due: Ch 30 - 3, 12, 17, 19, 26, 37, 38, 47, 51, 54, 57, 64, 72 Lecture Outline 1. The Biot-Savart Rule 2. Ampere's Law Now that we know the effects of magnetic fields on currents we can discuss the causes of magnetic fields. If currents feel the magnetic force, Newton's Third Law requires they must exert it as well. The purpose of this chapter is to learn to calculate magnetic fields caused by currents. The Biot-Savart Rule is a recipe that will always work. Ampere’s Law, while more fundamental, can be used to find fields only in cases where the current distribution is highly symmetric. 1. The Biot-Savart Rule

r dq dE = k 2 rˆ r r 2 2 dF = q dE r r 12 1 2 s + q I 1 r r 1 r dF = I sr ´ dB r r 12 1 2 dsr + I2 dq2

The electric force on a charge q1 is found from the field due to all the other charges. We begin by finding r dq 2 ˆ the field due to the point charge dq2, which is dE 2 = k 2 r . Next, the force on q1 due to dq2 is found r r r using definition of electric field, dF = q dE . Then integrate all the contributions from the rest of the 12 1 2 charge distribtion. The same proceedure would work for the magnetic force on I1 due to a distribution of r current I . The magnetic force on the current I in the magnetic field dB caused by a small current 2 1 2 r r element in I is, dF º I sr ´ dB which means that the magnetic equivalent of the electric charge element 2 12 1 2 dq is, q ® I rs ´ Þ dq ® Idsr´ Þ dq ® I dsr ´ . Using an analogy to the field due to the point charge 1 1 2 2 r dq r I dsr ´ r I dsr ´ rˆ dq , dE = k 2 rˆ ® dB = k 2 rˆ Þ B = k 2 . This is called the Biot-Savart Rule. 2 2 2 2 m 2 2 m ò 2 r r r

r r mo Ids ´ rˆ B = ò 2 The Biot-Savart Rule 4p r

m T ×m The magnetic constant is usually replaced with k = o º 10- 7 . m 4p A

30-1 Physics 4B Lecture Notes Example 1: Find the magnetic field due to a long straight wire carrying a current I. r r mo Ids ´ rˆ y Starting with the Biot-Savart Rule, B = ò 2 . 4p r r Note ds ´ rˆ = sinqdx and the field points out of the page. r m sinq a The magnitude of the field is,B = o I dx. 4p ò r2 q 2 x I is constant, r = acsc q and x = - acot q Þ dx = acsc qdq . r x I m I p sin q m I p m I m I ds B = o acsc2 qdq = o sinqdq = o ×2 = o 4p òq= 0 a2 csc2 q 4pa òq=0 4pa 2pa B At this point the field is out of the page. In general, the field I lines form rings around the wire. It is convenient to know the magnetic field due to a long straight wire.

m I B = o The Magnetic Field Due to a Long Straight Wire 2pr

Beyond the Mechanical Universe (vol. 35 Ch 8,9,12)

Example 2: Use the result of example 1 to find the force between two parallel currents. r r r Starting with the definition of magnetic field, F º I ´ B . I 2l 1 1 Since the direction is right, just find the magnitude, r F F = I B . Using the result of example 1, 2l 1 B1 moI1 mo I1I2 I2 F = I2l Þ F = × ×l just like before. 2pr 2p r Example 3: Find the magnetic field on the axis of a circular loop of radius a and magnetic dipole moment mr .

x

dsr q y r r r r a ds a dB q dq dsr z z q x

y r r r r mo Ids ´ rˆ moI ds ´ r Use the Biot-Savart Rule B = ò 2 = ò 3 . 4p r 4p r Notice that r = - acosqiˆ - asin qˆj + zkˆ and dsr = adq(- sin qiˆ + cosqˆj ).

30-2 Physics 4B Lecture Notes

iˆ ˆj kˆ dsr ´ r = - asinqdq acosqdq 0 - acosq - asinq z

dsr ´ r = [azcosqdqiˆ + azsin qdqˆj + a2 (sin2 q + cos2 q)dqkˆ ] 3 Since r3 = (a2 + z2 ) 2 and is constant, r m I 1 2p 2p 2p B = o × éaz ˆi cosqdq + azˆj sin qdq + a2kˆ dqù 3 ë ò0 ò0 ò0 û 4p (a2 + z2) 2

Doing the integrals, 2 r moI 1 2 ˆ r moI a ˆ B = × 3 [0 + 0 + a k 2p ] Þ B = × 3 k 4p (a2 + z2) 2 2 (a2 + z2 ) 2 r r mo m In terms of the dipole moment, B = × 3 . 2p (a2 + z2 ) 2

2. Ampere's Law Beyond the Mechanical Universe (vol. 35 Ch 19,20)

Ampere's Law states that the sum of the magnetic field along any closed path is proportional to the current that passes through. Mathematically, r r òB · ds = moienclosed Ampere's Law

y In order to understand this idea, consider a circular path around a wire that passes dsr through the center and is perpendicular to the plane of the circular path. We know m I r that the magnetic field due to the wire is, B = o and it is parallel to ds = rdq. 2pr I r r moI moI moI So, òB · ds = ò ×rdq = òdq = ×2p = moI consistent with Ampere's z 2pr 2p 2p Law.

y If we change the path so that it doesn't contain the current as shown on the right. The path needs to be broken up into four parts, r r r r r r r r r r r ds dsr òB · ds = òB · ds + òB · ds + òB · ds + òB · ds . x b y a dsr r r r r r r a b Along the x and y axes B^ ds Þ òB · ds = òB · ds = 0 . x y x I r m I ds Along b, B = o and it is parallel to ds = bdq so, 2pb B r m I m I m I p m I B · dsr = o ×bdq = o dq = o × = o . ò ò2pb 2p ò 2p 2 4 b b b

30-3 Physics 4B Lecture Notes

r m I m I m I p m I Similarly along a, B · dsr = - o ×adq = - o dq = - o × = - o . Putting it all together, ò ò2pa 2p ò 2p 2 4 a a a r m I m I òB · dsr = 0 + o + 0 - o = 0 also consistent with Ampere's Law. 4 4 Like Gauss's Law, Ampere's Law can be used to find the magnetic field of a sufficiently symmetric current distribution.

Example 4: Find the magnetic field inside and outside of a wire of radius a carrying a uniform current I. Sketch the field as a function of r the distance from the center. Outside the wire choose the circular path shown. Along this path the dsr symmetry requires B must be constant and it can only point along ds so, r òB · dsr = òBds = Bòds = B×2pr . a

The enclosed current is the entire current I. Applying Ampere's Law, r m I r o r òB · ds = moienclosed Þ B ×2pr = moI Þ B = . This is the same 2pr result as for an infinitely thin wire. Inside the wire choose the circular path shown. Along this path the r symmetry requires B must be constant and it can only point along ds so ds r again, r a òB · dsr = òBds = Bòds = B×2pr .

pr2 The enclosed current is i = ×I. Applying Ampere's Law, enclosed pa2 2 r pr m I ær B · dsr = m i Þ B ×2pr = m ×I Þ B = o × ö . The field grows linearly until the ò o enclosed o pa2 2pa è aø

surface of the wire is reached. The graph of B vs. r is shown below.

B moI 2pa

a r 1 a r

r a

30-4 Physics 4B Lecture Notes Example 5: Wire 1.00mm in diameter is wound around a 5.00cm diameter 1.00m long tube to create a solenoid. The wire carries a current of 100mA. Estimate the magnetic field in the center.

Apply Ampere's Law for the path B indicated. The current enclosed is 2 ienclosed = Ni where N is the number 1 3 of turns inside the path and i is the 4 B»0 current in the wire. The path integral l of the field can be broken into four parts, one for each side of the path. r r r r r òB · dsr = òB · dsr + òB · dsr + òB · dsr + òB · dsr . Along side 4 the field is close to zero. Along 1 2 3 4 r r r r sides 1 and 3 the field is perpendicular to the path so, òB · ds = 0 + òB · ds + 0 + 0 = Bl . 2 Assuming that the solenoid is so long that symmetry requires the magnetic field to be constant along the path. Using Ampere's Law, r N æ1000 B · dsr = m i Þ B = m Ni Þ B = m i = 4px10-7 ö( 0.100) = 1.26x10- 4 T ò o enclosed l o o ( )è 1.00 ø l N B = mo i The Magnetic Field Inside a Long Solenoid l

Example 6: The tube is bent into a circle to form a toroid. Find the magnetic field at the center.

The current enclosed by the indicated path is ienclosed = Ni where N is dsr the total number of turns and i is the current in the wire. The magnetic D field can only be a function of r and so it must be constant along the path r r r shown. So, òB · dsr = B ×2pr . Using Ampere's Law, i

r r N òB · ds = moienclosed Þ B2pr = moNi Þ B = mo i . 2pr Assume the inner circumference is equal to the length of the tube, r = r + D Þ 2pr = 2pr +pD = +pD. i 2 i l N æ 1000 ö The field is, B = m i = 4px10- 7 ç (0.100) = 1.09x10-4 T o +pD ( )è 1.00 +p(0.0500)ø l

30-5 Physics 4B Lecture Notes

Chapter 30 - Summary r r mo Ids ´ rˆ Biot - Savart Rule B = ò 2 4p r m I The Magnetic Field Due to a Long Straight Wire B = o 2pr r Ampere's Law B · dsr = m i ò o enclosed

N The Magnetic Field Inside a Long Solenoid B = mo i l

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