Sources of magnetic field Lots of details so far on how Magnetic Fields exert forces. But, where did those Magnetic Fields come from? B-fields are created by moving charges (currents). This seems quite opposite to what we might expect! Field created by a moving charge Magnetic Field created by a single moving charged particle at r r a position µ × ˆ q v r v = B v 0 π 2 4 r Constant µ 0 = 1.257 x 10 -6 Tesla meter/Amp. ˆ r r r is a vector in the direction of , but with magnitude = 1 ˆ r (often referred to as a unit vector). direction is from charge to where you are evaluating B !
Field created by a moving charge A new Right Hand Rule… µ × q v r ˆ v = B v 0 π 2 4 r Thumb in direction of the motion, and fingers then curl in direction of the B-field. Clicker Question µ × ˆ q v r v = B 0 v π 2 4 r What is the direction of the B- field at point P indicated? r A)Up and to the Left B)Down and to the Left C)Out of the Page Point P D)Into the Page E)None of the Above
Example: Forces between two moving protons 2 1 e = F E r y ˆ πε 2 4 r 0 µ × µ ˆ e v r ev v = = B z ˆ v 0 0 π π 2 2 4 r 4 r = − × = F B r e ( v ) B r r µ ev ⎛ ⎞ − × = e ( v x ˆ ) 0 z ˆ ⎜ ⎟ π 2 4 r ⎝ ⎠ µ µ 2 2 ev e v ( ) − × = ˆ ˆ e ( v ) 0 x z ˆ 0 y π π 4 r 2 4 r 2 Magnetic field of a current element Now the B-field created by many moving charges (i.e. current flowing in a wire). µ × ˆ dQ v r r = = d B 0 d v π 2 4 r µ × ˆ v r I r = 0 ( nqAdl ) d π 2 4 r dl µ × r ˆ I d l r 0 π 2 r 4 r X dB Biot-Savart first discovered this experimentally.
Magnetic field of a current element µ × I d l r r ˆ = d B 0 v π 2 4 r I Magnetic field of a current-carrying wire µ × I d L r ˆ v = d B 0 v π 2 4 r µ × ˆ I d L r v = = B tot d B v v 0 ∫ ∫ π 2 4 r This can be a very difficult integral to evaluate.
Magnetic field from an infinite straight wire µ × I d L r ˆ v = d B 0 v π 2 4 r µ I dy = θ d B 0 sin ( into the page) v π 2 4 r = + 2 2 2 r y x Now a little geometry θ = = 2 + 2 sin x / r x / x y µ I xdy = d B 0 v ( ) 3 / 2 π 4 2 + 2 x y Magnetic field from an infinite straight wire (cont.) µ I xdy = d B 0 v ( ) π 3 / 2 4 + 2 2 x y +∞ µ I xdy = = B d B 0 v v ∫ ∫ π + 2 2 3 / 2 4 ( x y ) − ∞ µ I = B 0 (into the page) v π 2 x
Magnetic field from an infinite straight wire (cont.) This is a key result! µ I B-field a perpendicular distance = | B v | 0 x away from an infinite (or very π 2 R long) wire. Clicker Question A long wire has a current moving as shown. What is the direction of the B-field created by the wire just below the wire? L i r A)Into the Page B)Out of the Page B=? C)To the right µ × i d L r ˆ v = d B 0 v D)Down π 2 4 r E)None of the Above
Clicker Question A long wire has a current moving as shown. What is the direction of the B-field created by the wire just above the wire? B=? r L A)Into the Page i B)Out of the Page C)To the right µ × ˆ i d L r v = d B v 0 D)Down π 2 4 r E)None of the Above Example with numbers Power line has 500 Amps going through it. What is the B-field strength 15 meters below on the ground? µ × -6 I (1.26 10 Tm / A )( 500 A ) = = = × − 6 B 0 6 . 7 10 T π π 2 R 2 ( 15 m )
Interaction between two current carrying wires The two wires may exert forces on each other through Magnetic Interactions. I I’ • Think of the Red Wire as creating a B-field. • Then think of that B-field creating a force on the moving charges (current) in the Blue Wire . Interaction between two current carrying wires = ' × F I L B v v v µ I = ˆ B r 0 r π 2 r µ II ' L = = F I ' LB 0 π 2 r
Clicker Question What is the direction of the Force I I acting on the Blue Wire ? = × F I L B v v v A)Up B)Right C)Left D)Into the Page E)Out of the Page The B-field from the Red Wire at the location of the Blue Wire is into the i 1 i 2 page. µ i = B 0 1 ( into the page) v 1 π 2 R Then the Force on the Blue Wire is to X the left. µ i = × = F i L B i L 0 1 ( left) v v v 2 2 1 2 π 2 R R
Clicker Question What is the direction of the Force i i acting on the Red Wire ? = × F I L B v v v A)Up B)Right C)Left D)Into the Page E)Out of the Page The B-field from the Blue Wire at the location of the Red Wire is out of the i 1 i 2 page. µ i = B 0 2 ( out of the page) v 2 π 2 R Then the Force on the Red Wire is to . the right. µ i = × = F i L B i L 0 2 ( right) v v v 1 1 2 1 π 2 R R
Interaction between current-carrying wires i i Wires with parallel currents attract each other. i What happens if we flip the direction of one current? Wires with antiparallel currents repel each other. Try following the procedure we just outlined to confirm this for yourselves. i Field through a circular loop What about the B-field at the center of a circular loop of wire of radius a and current I ? µ µ I dl I dl = = dB 0 0 π 2 π 2 + 2 4 r 4 ( x a ) µ I adl = θ = dB x dB cos 0 π 2 + 2 3 / 2 4 ( x a ) µ µ I adl Ia = = = B dB 0 0 dl x ∫ x ∫ ∫ π + π + 2 2 3 / 2 2 2 3 / 2 4 ( x a ) 4 ( x a ) µ 2 Ia = B x 0 + 2 2 3 / 2 2 ( x a )
Field through a circular loop Clicker Question Which point A or B has the larger magnitude Magnetic Field? I I µ × r ˆ I d l r = d B 0 v π 2 4 r B A A B C : The B-field is the same at A and B. Answer: Case B has the larger magnetic field. Use the Biot-Savart Law to get the directions of the B-field due to the two semi-circular portions of the loop. In A the two fields oppose each other; in B they add.
Clicker Question = µ 2 Ia In the limit as x >> a, the expression B 0 for the B-field becomes? 2 + 2 3 / 2 2 ( x a ) µ Ia 2 = A) B 0 B=? 3 2 x = µ 2 Ia B) x B 0 2 + 2 2 ( x a ) µ I a = B 0 C) 2 a I Magnetic Dipole moment µ 2 Ia = B 0 x ˆ v 3 2 x B=? µ π 2 µ I a IA = = B 0 x ˆ 0 x ˆ * A = area of loop π π 3 3 2 x 2 x x v = µ I A v Magnetic Dipole Moment a v = µ µ v I B 0 π 2 x 3
Magnetic Dipole moment Any current loop looks like a Magnetic Dipole far away. µ µ v = B 0 v π 3 2 x The B-field drops as the distance^3 And depends on the Magnetic Dipole Moment. v = µ I A v Clicker Question A square loop of side length a of wire carrying current I is in a uniform magnetic field B. The loop is perpendicular to B (B out of the page). What is the magnitude of the net force on the wire? = × F I L B v v v A: IaB B B: 4IaB C: 2IaB D: 0 E: None of these I
Clicker Question The same loop is now in a non-uniform field. r = ⋅ B Bz $ , where B = B(y) = A y where A is a constant. The direction of the net force is? B stronger B y A C D B weaker x E: net force is zero Uniform fields produce zero net force!!! In a uniform B-field, regardless of the orientation between the B and the Magnetic Moment of the loop µ , the net force is always zero. However, that does not mean the net torque is zero!
Magnetic torque If µ is not parallel to B, then there is µ a net torque. τ = ∑ × r F v v v τ = × × = θ r ( I L v B v ) 2 [( L / 2 ) I LB sin ] v v ∑ τ = 2 B θ IL sin v µ τ = × I A v B v v τ = µ × B v v v The DC motor τ = µ × Torque wants to twist the B v v v loop so that µ and B align.
Clicker Question Two loops of wire have current going around in the same direction. i 2 The forces between the loops is: A)Attractive i 1 B)Repulsive C)Net force is zero. Magnetic Field Lines and Flux
Gauss’s Law for Magnetic Fields Magnetic monopoles (so far) do not exist!!! There are no sources of magnetic flux ⋅ = =>Flux through a r r B d A 0 closed surface: ∫ Ampere’s law Is there something like Gauss’ Law for Electric Fields that helps us solve for Magnetic Fields in a simpler way for cases with nice symmetries? ⋅ = µ B d l v I v ∫ 0 thru loop For any closed imaginary loop where the current is constant, the above relation is true.
Clicker Question ⋅ = µ B d l v I v ∫ 0 thru loop We need a sign convention for I(thru). Place the fingers of your right hand around the imaginary loop and your thumb points in the direction of positive I(thru). What is I(thru) in the below case where all three wires have 5 A? A) I(thru) = +15 Amps X B) I(thru) = +5 Amps X C) I(thru) = -5 Amps D) None of the above Ampere’s law for a straight wire i Place an imaginary circular loop of radius R around the wire. ⋅ B d l v v ∫ loop Vector dot product!
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