Thursday, April , 1999
Announcements:

Lecture notes:

• Example of Biot and Savart Law
• 
• Find the electric field due to current in a straight wire of infinite length
• What is B at point P?
• Take point O to be the orgin. The positive z axis is out of the screen.
• i ds and r are on the x -y plane. dB is in the +z direction.
• dB = (m_o/4p)( i ds(cross)r / r²)
• Absolute value of dB =  (m_o/4p)( i ds / r²)sin q
• We now need to express the "sum" of the absolute value of dB in terms of a and q₁ and q₂.
• absolute value of ds = dx
• x = a cot (180 - q) = - a cotq.
• dx = a csc²q dq.
• r² = a² + x² = a² + a² cot²q = a²( 1+ cot²q ) = a² csc²q .
• Therefore, Absolute value of dB =  (m_o/4p)( i a csc²q dq sinq / a² csc²q)
• Simplified, Absolute value of dB =  (m_o/4p)( i sinq / a) dq
• B = (m_oi/4pa)I (integral from q₁ to q₂) sinq  dq  = -(m_oi/4pa) cosq (Evaluated from q₁ to q₂)
• B = (m_oi/4pa)(cos q₁ - cos q₂)
• If the wire is infinitely long, q₁ goes to 0 and q₂ goes to p
• So for an infinitely long wire, B = (m_oi/4pa)(2) = (m_oi/2pa)
• What is the direction of B due to the straight wire?
• Use right hand rule, Point thumb of right hand along the direction of i, the fingers will indicate the direction of B
• Above the wire the field is coming out of the screen, Below the wire, the field is going into the screen.
• See also Problem 30 -9 and 30 -14

• The distance between the wires is a, the length of the wires is l.
• What is the B due to wire 2 at 1?
• B₂ =  (m_oi₂/2pa)
• The force due to B₂ on a section of wire 1 (total length l) is F₁ = i₁ l B₂ =  (m_oi₁ i₂ l/2pa)
• Force per unit length F₁ / l =  (m_oi₁ i₂ l/2pa)
• What is the direction ?
• B₂ at i₁ is out of the screen, therefore F = i (l cross B)
• F1 is towards i2
• Similarly, the force on i2 due to i1 is towards i1
• Parallel conductor carrying current in the same direction attract each other

• In electrostatics, we can calculate, in priniciple, the electric field if we know the distribution of the elelectric charges. But it could be messy and tedious. If the situation has a high degree of symmetry, we can make use of Gauss' Law. Then the solution is easy.
• Similarly, we can use Biot-Savart Law to calculate the magnetic field at any point in space due to any current distribution, but it may be messy and difficult. For problem of high symmetry, we can use Ampere's Law.
• I(integral) B (dot product) ds = m_oi_enclosed
• 
• Now, I(integral) B (dot product) ds = m_o(i1 + i2 + i3)
• If the loop has a complicated shape, Ampere's law is not helpful in calculating B at a particular point.
• Example using Ampere's Law
• Megnetic Field due to a long wire
• If we want to  know the value of B at some point P , draw an amperian loop, symmetrical around i and passes through P.
• 
• then, I(integral) B (dot product) ds = m_oi_enclosed
• B is parallel to ds, so B (dot product) ds = Bds
• The loop is symmetrical , so B is the same along the whole loop
• Therefore, I(integral) B (dot product) ds =  B I(integral) ds = B2pr = m_oi_enclosed
• B =  m_oi_enclosed /2pr
• Magnetic due to a Toriodal Coil ( doughnut)
• See figure 30-21 from the text for a picture of a toriod
• B outside and inside the toriod = 0
• B within the coils of the toriod does not equal zero.
• What is the magnitude of B?
• Draw a circular amperian loop of radius r. This loop contains N turns so...
•  I(integral) B (dot product) ds = (m_oi_enclosed)N
• Simplifies to m_oi_enclosed N/2pr