ELEC 3105 Basic EM and Power Engineering Biot-Savard Law Applications of Biot-Savard Law

Biot-Savard Law Usually we deal with current localized to wires rather than spread over space. When the current is contained by thin wires, “we can use the magnetic vector potential to derive an integral from which the magnetic field can be found directly from the wire location and currents”. I P Origin

Biot-Savard Law KISS Rule We will find the magnetic field at some point P located at vector from the origin due to a short segment of wire located at from the origin, carrying current I, where the current is in the direction of the vector , as shown below. find here P I Origin small segment of wire at I

Biot-Savard Law This expression is The Biot-Savard Law Consider a small segment of wire of overall length P Same result as postulate 2 for the magnetic field I

Biot-Savard Law This expression is The Biot-Savard Law Consider a small segment of wire of overall length The magnetic field produced by all the short current elements along the line is the Biot-Savard law. P An integral is another way of saying “principle of superposition” applies to magnetic fields. I

Biot-Savard Law & Current Density J (A/m2) find here P I Origin 𝑅 I The current in the differential volume element dxdydz is:

Biot-Savard Law & Surface Current Density K (A/m) find here 𝐾 P I Origin 𝑅 dA I The current in the differential surface element dA is:

ELEC 3105 Basic EM and Power Engineering START Applications of Biot-Savard Law BIOT from RAMA

Magnetic field produced by a circular current ring

Magnetic field produced by a circular current ring We can use the Biot-Savard Law We will first consider the magnetic field present at the center of a current carrying ring. y x

Magnetic field produced by a circular current ring All contributions from each line segment point in the same direction: in the direction of the axis of the ring. y x I out of page Side view Top view I into page

Magnetic field produced by a circular current ring Side view Top view Starting from the Biot-Savard Law for a single segment. x We get:

Magnetic field produced by a circular current ring Side view Top view x Direction: As shown by arrows in side view

Magnetic field produced by a circular current ring Summation over all segments around the ring y Top view x Magnetic field at the center of the circular current ring

Magnetic field produced by a circular current ring x y 3-D view z We will now consider the magnetic field present on the central axis of a current carrying ring. We can use the Biot-Savard Law

Magnetic field produced by a circular current ring All contributions from each line segment point in different directions: We must consider components of . z 3-D view z y x I out of page I into page

Magnetic field produced by a circular current ring z Horizontal components will cancel in pairs around the segments of the current ring I out of page I into page Z components will add around the segments of the current ring

Magnetic field produced by a circular current ring Horizontal components will cancel in pairs around the segments of the current ring Z components will add around the segments of the current ring

Magnetic field produced by a circular current ring 3-D view I out of page I into page z Can use Biot-Savard Law We need only concern ourselves with the magnitude of since we know the final direction of is along the z direction.

Magnetic field produced by a circular current ring 3-D view I out of page I into page z Expression for magnetic field produced by one current segment of the ring.

Magnetic field produced by a circular current ring I out of page I into page z 3-D view z y x Summation around ring: Magnetic field on the central axis of the circular current ring

Magnetic field produced by a circular current ring We only considered the central axis field The full field is obtained when we will discuss the magnetic dipole magnetotherapy

Magnetic field of a long FINITE solenoid

Magnetic field of a FINITE solenoid Current out of page Axis of solenoid P L Current into page finite coil of wire carrying a current I Evaluate B field here Radius of solenoid is a. Cross-section cut through solenoid axis

Magnetic field of a FINITE solenoid Segment of the solenoid coil arc length 1 2 3 4 5 Current out of page Axis of solenoid

Magnetic field of a FINITE solenoid The solenoid consists on N turns of wire ”rings of current” each carrying a current I. The number of turns per unit length of solenoid can be expressed as:

Magnetic field of a FINITE solenoid The current in a segment (dI) can be expressed as the current in one ring (I) multiplied by the number of turns per unit length (N/L) and multiplied by the segment length .

Return to this slide for a moment z 3-D view y x Each ring of the solenoid contributes a magnetic field. Can use this expression for our finite solenoid. Sum over the rings. Magnetic field on the central axis of the circular current ring

Magnetic field of a FINITE solenoid Segment of the solenoid coil arc length 1 2 3 4 5 Current out of page Axis of solenoid

Magnetic field of a FINITE solenoid sub in 1 2 3 4 5

Magnetic field of a FINITE solenoid z L We can now sum (integrate) the expression for over the angular extent of the coil. I.e. sum over all the rings of the finite length solenoid.

Magnetic field of a INFINITE solenoid z L INFINITE SOLENOID RESULT infinite solenoid result

Magnetic field at one end of a FINITE solenoid z L Magnetic field is ½ that of center

Magnetic flux Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 Goes by the name of magnetic flux density vector 𝐵

Magnetic flux: CLOSED SURFACE Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 =0 Magnetic field lines loop back on themselves No magnetic source No magnetic sink No magnetic monopole No magnetic charge

Divergence of a Magnetic Field For a closed surface Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 =0 No magnetic monopole No magnetic charge Divergence theorem 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝛻 ∙ 𝐵 𝑑𝑣𝑜𝑙 𝛻 ∙ 𝐵 =0 𝛻 ∙ 𝐷 = 𝜌 𝑣 𝛻 ∙ 𝐻 =0 Electric charge exist

Magnetic flux Given that the radial magnetic field is 𝐵 = 𝜇 𝑜 2.39× 10 6 𝑐𝑜𝑠 𝜙 𝑟 𝑟 , calculate the magnetic flux Φ 𝑚 through the surface defined by − 𝜋 4 ≤𝜙≤+ 𝜋 4 and 0≤𝑧≤1. We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates. Solution Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 Remainder of solution presented in class

Magnetic flux In cylindrical coordinates 𝐵 = 2.0 𝑟 𝜙 , calculate the magnetic flux Φ 𝑚 through the surface defined by 0.5≤𝑟≤2.5 and 0≤𝑧≤2. We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates. Solution Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 Remainder of solution presented in class