1 Chapter 29: Magnetic Fields due to Currents Introduction What are we going to talk about in chapter 30: How do we calculate magnetic fields for any distribution of currents? What is the magnetic field due to a straight long wire carrying a current? What is the magnetic field due to a loop carrying a current. How much is the force between two parallel current carrying wires? What is Ampere ’ s law about? What is the field inside a solenoid? What is a magnetic dipole? What is the torque on a current loop?

2 Biot-Savart law: dB = (  o /4  ) i dsxř/r 2  o = 4  T m/A 29.2: Calculating the magnetic field due to a current: BSL is an inverse square law!! If ds and ř are parallel, the contribution is zero!

3 Special cases: For a straight wire making angles  1 and  2 : B = (    i/4  a)(cos  1 - cos  2 ) grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element. The direction is found through the right hand rule: Prove it!

4 For infinite straight wire: B = (  o i/2  a) At the center of a circular arc with angle  : The radial part extending from the arc does not contribute; why? At the center of a circular loop: B = (  o i/2R) Checkpoint #1 B = (  o i  /4  R)

5 B a =  o i a /(2  d) Therefore, the force felt at wire #b is: 29.3: Force between two parallel currents: For two long, straight parallel wires a distance [d] apart, the magnetic field created by current ia at wire #b is: F ba = I b L x B a L and B are perpendicular, therefore F ba = i b L B a F ba /L =  o i a i b /(2  d)

6 What about the force which wire #a feels? The same!! Why? F ab /L =  o i a i b /(2  d) In which direction are the forces? Towards one another if the currents are parallel and away from one another if they are anti-parallel. In this way we can define the ampere!!

7 The rail gun: Checkpoint #2 Is a device in which a magnetic force can accelerate (~10 6 g) a projectile to a high speed (~10 km/s) in a short time (~ 1 ms).

8 29-4: Ampere’s law: Ampere ’ s law states that: the closed path line integral of B. ds around a circle concentric with the current equals  o i enc. As we learned in section 30.1, if a current carrying wire is grasped with the right-hand with the thumb in the direction of the current, the fingers curl in the direction of B.

9 Special cases: The magnetic field outside a long (thick/ circular) straight wire: The magnetic field inside a long (thick/ circular) straight wire with uniform current density: B =  o i r/(2  R 2 ) for r < R B =  o i/(2  r) for r > R

10 Checkpoint #3 B =  o J s /2 What is the magnetic field created by an infinite uniform current sheet J s, with a current i over a perpendicular length of the sheet L such: i = J s /L? Interaction: Prove it!

: Solenoids and Toroids: The field inside the solenoid is ~ uniform. The field between the turns tend to cancel. The field outside the solenoid is weak! The field of a solenoid is similar to that of a bar magnet! An ideal solenoid is one for which the turns are closely spaced and the length is long compared to the radius.

12 Applying Ampere ’ s law to an ideal solenoid gives: So, now we know how to create strong uniform magnetic fields!! Why do we use superconducting coils? B =  o (N/L) I =  o n Iinside B= 0outside

13 What is the magnetic field created by a toroid? Note that B is not everywhere constant inside the toroid!! B =  o N i/(2  r)inside B = 0outside

: A current carrying coil as a magnetic dipole: B(z) = (  o /2) i R 2 /(R 2 +z 2 ) 3/2 We have already seen that if a circuit of magnetic dipole  is situated in a magnetic field B, the circuit experiences a torque  produced such that:  =  x B Moreover, one can show that for points on the central axis (take it to be the z-axis) of a single circular loop, a circulating current [i] produces a magnetic field:

15 Notice the similarity with the electric field and the electric dipole!! Checkpoint #4 The circular loop current acts like a magnet: B(z) = (  o /2  )  /z 3 For points far from the loop (still on the z-axiz), this can be cast in the form: