prerequisites for magnetism before you start to study
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Prerequisites for Magnetism Before you start to study electromagnetism you must review these topics: Cross product of two vectors: modulus and direction (right-hand rule and screw rule). Torque of a vector (force). Surface vector


  1. Prerequisites for Magnetism  Before you start to study electromagnetism you must review these topics:  Cross product of two vectors: modulus and direction (right-hand rule and screw rule).  Torque of a vector (force).  Surface vector of a flat surface.

  2. Unit 6: Magnetic forces  Introduction. Magnetic Field.  Forces exerted by a magnetic field:  Force on a moving electric charge.  Application: Hall effect.  Force on a conductor flowed by a current.  Effect of a uniform magnetic field on a flat current- carrying loop. Magnetic moment. Electric motor.

  3. Introduction. Magnetic field Magnets: Poles attract or repel each other in the  same way as electric charges. Poles of a magnet: north pole and south pole  ( similar to + and – charges) . Differences Magnetic – Electric poles: A magnetic pole can’t be isolated.  Magnetic monopoles don’t exist. If you break a magnet in two pieces, both pieces  have their north and south poles. Where the effects of magnetic poles appear, this is said to be a Magnetic Field (B). A magnetic field can be represented as a vector B at each point in the field. Field lines are parallel to B, as in an electric field. Tipler, chapter 26

  4. Force on a moving charge in a magnetic field  When an electric charge q is moving with velocity v, inside a magnetic field B, a force F acts on the charge, given by � � � (empirical result) : F = q v × B F B B Right-hand rule - α q α or q v + v screw rule F If v and B are parallel F=0  Unit I.S: Tesla. [ F ] [ ]  Other unit used: Gauss 1 G=10 -4 T. − 2 − 1 B = = MT I  Earth field: 0,5 × 10 -4 T . q v [ ] [ ]  Magnetic resonance: 2~5 T . Tipler, chapter 26, section 26.1

  5. Application: Hall effect It occurs when a current is flowing through a conductor inside a magnetic field perpendicular to I. � So, a magnetic force ( ) acts on charge carriers. Let’s F m suppose such carriers are electrons (-). I � � V d is drift speed � F = q v × B v d m d F m J � � � B F = v B = q qv B m d d This magnetic force separates positive and negative charges perpendicularly to the electric current and magnetic field. Tipler, chapter 26, section 26.4

  6. Application: Hall effect When charges are split up, an electric field appears (and so a � drop in potential), producing an electric force ( ) opposite � � F e to the magnetic force: F e = q E I b V H F e E v d a J F m B When equilibrium is reached, magnetic and electric forces are   balanced: F = F qv B = qE E = v B m e d d Drop in potential V H is Hall voltage (order of μV): I IB V = Ea = v Ba = Ba = H d nqS nqb

  7. Application: Hall effect The type of charge carriers determine the polarity of V H (useful to distinguish n or p semiconductor type). I V H v a Negative charge carriers F J I B -V H ⊕ ⊕ ⊕ ⊕ v a Positive charge carriers F J B It is also useful to build speed selectors. Only those charges with that speed cancelling electric and magnetic forces will cross the conductor without deflection.

  8. Force on a conductor with current inside a B When a conductor with current flowing through it is inside a magnetic  field, the force on the conductor is the sum of the forces acting on all the charges in the conductor: Taking an infinitesimal piece of conductor (lenght dl), the total number  n S dl of charge carriers in dl will be: n= density of charge carriers I = JS = nqv S d � � � � � ( ) And the force acting over dl: F = v × B = l × B d q n S dl I d d BE CAREFUL: dl is a vector in the same direction than v d and I � is called a current element l Id Tipler, chapter 26, section 26.1

  9. Force on a conductor with current  If we consider a conductor with any shape, the total force on the conductor will be the sum of all the forces (integration) on it: l d B a b � � � � �  b  b ( ) F = l × B = l × B I d I d a a

  10. Example: Force on a straight conductor in a uniform B  Let’s take a straight conductor inside a uniform magnetic � field B: B � d l I a � � b l F � � � � � � � b b   ( ) F = I d l × B = I d l × B = I l × B a a l is a vector from a to b, in the I and B are constants same direction than the current. from a to b

  11. Effect of a uniform magnetic field on a flat current- carrying loop. Magnetic moment.  Let’s take a flat loop inside a uniform B, with a current I: I � B b a Tipler, chapter 26, section 26.3

  12. Effect of a uniform magnetic field on a flat current- carrying loop. Magnetic moment.  On each side of the loop, a force will act: � � � � � F F = I b × B 2 F 1 � � � � � 3 ( ) F = I − b × B = − F b 2 1 � � � � � � ( ) F = I − a × B F 3 F 1 � � � B � 4 F = I a × B = − F 4 3 � a =  � � 4  Resulting force on loop is zero: F F = 0 i i = 1

  13. Effect of a uniform magnetic field on a flat current- carrying loop. Magnetic moment.  Torque of each force in relation to point O (loop centre): � � � a τ � = − × F � 1 1 2 F � 2 F � � � 3 a τ = × F O b 2 2 2 � � � � � F � B b F 1 τ = − × F 4 3 3 2 � � � � b a τ = × F 4 4 2

  14. Effect of a uniform magnetic field on a flat current- carrying loop. Magnetic moment.  Resulting torque about O is: =  � � 4 � � � ( ) τ τ = I b × a × B i i = 1 � � � surface vector of the loop (direction given by S = b × a current in loop according to right-hand rule) � � magnetic moment of the loop [ ] μ 2 µ = = I S IL  So, the resulting torque in the loop can be written as: � � � τ μ = × B

  15. Effect of a uniform magnetic field on a flat current- carrying loop. Magnetic moment.  So, when we apply an I to a flat loop in a uniform B, the loop turns until magnetic moment µ is aligned with B ( τ =0), the loop remaining in equilibrium. I I � � � � � S μ = I S � B B S � � μ = I S  If there isn’t just one loop, but many (N) loops (a coil): � � μ = S NI

  16. Application: electric engine An electric motor consists of a set of loops carrying a current inside a magnetic field. Its behaviour is that of a receptor. Features: ε ’ and r’ � � � � � B iN S B τ = µ ∧ = ∧ ε ’ and ω are   P ' i kiNSB ' i ' k ' = τ ω = ε ω = ε ε = ω directly related m

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