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PHYSICS 1B Today’s lecture: The Biot-Savart Law and Ampere’s Law Electricity & Magnetism 22/08/2013 PHYSICS 1B – Sources of the Magnetic Field The Biot-Savart Law - Introduction Jean-Baptiste Biot and Félix Savart conducted experiments on the force exerted by an electric current on a nearby magnet. They arrived at a mathematical expression that gives the magnetic field at some point in space due to a current. The magnetic field described by the Biot-Savert Law is the field due to a given current carrying conductor. Be careful not to confuse this field with any external field applied to the conductor from some other source! Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field The Biot-Savart Law – Set-Up Biot and Savart measured the magnetic field, dB, at some distance r from a wire, which was carrying a steady current of I. The length element for a small section of the wire is ds. They found that dB is: • proportional to the current, I, the length, ds and sin θ • inversely proportional to r2 Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field The Biot-Savart Law – Set-Up They found that dB is proportional to the current, I, the length, ds and sin θ, and inversely proportional to r2 These observations are brought together in the Biot-Savart law: 𝜇0 𝐼 𝑑𝒔 × 𝒓 𝑑𝑩 = 4𝜋 𝑟2 This introduces a new constant, μ0, which is called the permeability of free space. T. m −7 𝜇0 = 4𝜋 × 10 A The magnetic field is due to the current-carrying conductor. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Total Magnetic Field dB is the field created by the current in the length segment ds. To find the total field, B, we have to sum up all the contributions from all the current elements I ds . Therefore: 𝜇0 𝑩= 4𝜋 𝐼 𝑑𝒔 × 𝒓 𝑟2 which is the integral over the entire current distribution. The Biot-Savart Law is also valid for a current consisting of charges flowing through space (i.e. we could apply it to the particle beam in an accelerator). Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Quick Quiz Consider the current in the length of wire shown below. Rank the points A, B, and C in terms of the magnitude of the magnetic field due to the current in the length element, ds, shown, from the greatest to the least. a) A, B, C b) B, C, A c) C, B, A d) C, A, B e) An equal field applies at all these points Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Quick Answer Consider the current in the length of wire shown below. Rank the points A, B, and C in terms of the magnitude of the magnetic field due to the current in the length element, ds, shown, from the greatest to the least. b) B, C, A Point B is the closest to the current element. Point C is farther away, and in addition, the field is reduced by the sin θ factor in the cross product 𝑑𝒔 × 𝒓. The field at A is zero because θ = 0. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field Compared to Electric Field Distance The magnitude of the magnetic field varies as the inverse square of the distance from the source. The electric field due to a point charge also varies as the inverse square of the distance from the charge. Direction The electric field created by a point charge is radial in direction. The magnetic field created by a current element is perpendicular to both the length element ds and the unit vector 𝒓. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field Compared to Electric Field Source An electric field is established by an isolated electric charge. The current element that produces a magnetic field must be a part of an extended current distribution, and so you therefore have to integrate over the entire distribution of current to work out the magnetic field. End of Field Lines Magnetic field lines have no beginning and no end – they instead form continuous circles. Electric field lines begin on positive charges and end on negative charges. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field for a Long, Straight Conductor First, find the field contribution from a small element of the current, then integrate over the current distribution. We have thin, straight wire carrying a constant current, I, so therefore: 𝑑𝒔 × 𝒓 = 𝑑𝑥 cos 𝜃 𝒌 So now, we just integrate over the current elements from one end of the wire to the other… Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field for a Long, Straight Conductor So now, we just integrate over the current elements from one end of the wire to the other… 𝜇0 𝐼 𝑩=− 4𝜋𝑎 ∴ 𝜃2 cos 𝜃 𝑑𝜃 𝜃1 𝜇0 𝐼 𝑩= sin 𝜃1 − sin 𝜃2 4𝜋𝑎 This is example 30.1 in the text book… Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field for a Long, Straight Conductor Special Case 𝜇0 𝐼 𝑩= sin 𝜃1 − sin 𝜃2 4𝜋𝑎 If the conductor is an infinitely long, straight wire, then 𝜃1 = 𝜋/2 and 𝜃2 = −𝜋/2. The field then simplifies to become: 𝜇0 𝐼 𝑩= 2𝜋𝐴 Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field for a Long, Straight Conductor Direction The magnetic field lines are circles concentric with the wire. The field lines lie in planes perpendicular to the wire. Magnitude of B is constant on any circle of radius a. The right-hand rule to determine the direction of B is shown. Simply grasp the wire with the thumb in the direction of the current. Your fingers wrap in the direction of the field! Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field for a Curved Wire Segment Find the field at point O due to the curved wire segment. I and a are constants. ds is parallel to r along the line A to A’ and along CC’, so we only need to work out the field resulting from AC. We find that 𝑩= 𝜇0 𝐼 𝜃 4𝜋𝑎 , where θ is in radians… See example 30.2 for the working here… Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field for a Circular Wire Loop We can use our previous result, namely: 𝜇0 𝐼 𝑩= 𝜃 4𝜋𝑎 to find the field for a circular loop of wire. For a circular loop of wire, 𝜃 = 2𝜋, and therefore: 𝜇0 𝐼 𝜇0 𝐼 𝑩= 2𝜋 = 4𝜋𝑎 2𝑎 This is the field at the centre of the loop, which has radius a. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Field Lines for a Loop The left figure, (a), shows the magnetic field lines surrounding a current loop. The right figure, (b), shows the field lines around a bar magnet. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Force Between Two Parallel Conductors In the figure to the right, there are two parallel wires carrying steady currents I1 and I2 . The field B2 , which is the result of the current in wire 2, exerts a force on wire 1 of 𝐹1 = 𝐵2 𝐼1 𝑙 𝜇 𝐼 We know that 𝐵2 = 0 2 , from earlier – so we can 2𝜋𝑎 substitute this to get: 𝜇0 𝐼1 𝐼2 𝑙 𝐹1 = 2𝜋𝑎 Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Force Between Two Parallel Conductors 𝜇0 𝐼1 𝐼2 𝑙 𝐹1 = 2𝜋𝑎 This equation tells us that: Parallel conductors that carry current in the same direction attract one another. Parallel conductors that carry current in opposing directions repel one another. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Magnetic Force Between Two Parallel Conductors This result is often expressed as the magnetic force between two wires, FB. It can be rearranged to give us the force per unit length: 𝐹𝐵 𝜇0 𝐼1 𝐼2 = 𝑙 2𝜋𝑎 The derivation here assumes that both wires are long compared to their separation distance, although technically only one wire needs to be long. The equations accurately describe the forces exerted on each other by a long wire and a straight, parallel wire of limited length, 𝑙. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Quick Quiz If, in the figure to the right, we set I1 = 2 A and I2 = 6 A, which is true? a) F1 = 3 F2 b) F1 = F2 / 3 c) F1 = F2 Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Quick Answer If, in the figure to the right, we set I1 = 2 A and I2 = 6 A, which is true? c) F1 = F2 Newton’s third law requires that F1 = F2. Another route to the same answer is to realise that: 𝜇0 𝐼1 𝐼2 𝑙 𝐹1 = 2𝜋𝑎 Gives the same result whether the multiplication is 2x6 or 6x2. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Quick Quiz A loose spiral spring is hung from the ceiling, and a large current is passed through it. The coils move: a) Closer together b) Farther apart c) They do not move at all Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Quick Answer A loose spiral spring is hung from the ceiling, and a large current is passed through it. The coils move: a) Closer together The coils act like wires carrying parallel currents in the same direction, and hence they attract one another. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Definition of the Ampere Earlier, we introduced the concept of Electric Current, and mentioned the unit is the ampere (A). The force between two parallel wires can be used to define the ampere. When the magnitude of the force per unit length between two long, parallel N wires that carry identical currents and are separated by 1m is exactly 2 × 10−7 m then the current that is carried by each wire is defined to be 1 A. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Sources of the Magnetic Field Definition of the Coulomb The SI unit of charge, the coulomb, is actually defined in terms of the ampere. When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C. Electricity & Magnetism – Magnetism 21/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Andrew-Marie Ampère 1775 – 1836 French physicist and mathematician Carried out experiments following Oersted’s observation that a compass is deflected by an electric current. Showed that electricity and magnetism were linked, and so is often credited as being the discoverer of Electromagnetism. Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields The Magnetic Field for a Long, Straight Conductor As we saw earlier, the magnetic field lines are circles concentric with the wire. The magnitude of the field is constant on any circle of radius a. The right-hand rule for determining the field’s direction is shown to the right… Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields The Magnetic Field of a Wire We can use a compass to detect the magnetic field, reproducing Oersted’s experiment. When no current flows in the wire, there is no field due to the current. The needle will therefore point roughly towards the Earth’s north geographic pole. Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields The Magnetic Field of a Wire When the wire carries a current, a compass needle deflects in a direction tangent to the circle. This shows the direction of the magnetic field produced by the wire. Just like our old friend, the iron filings! Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Ampere’s Law The line integral of B.ds around any closed path equals μ0I, where I is the total steady current passing through any surface bounded by the closed path. i.e. 𝑩. 𝑑𝒔 = 𝜇0 𝐼 Ampere’s law describes the creation of magnetic fields by all continuous current configurations – and is particularly useful when the current configuration has a high degree of symmetry. Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Ampere’s Law – which way to integrate? 𝑩. 𝑑𝒔 = 𝜇0 𝐼 In order to evaluate the direction you should integrate around a given amperian loop (i.e. a closed path carrying current): Put the thumb of your right hand in the direction of the current through the loop, and your fingers will then curl in the direction you should integrate in. For more discussion on the background to Ampere’s law, see pages 869-870! Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Quick Quiz Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed paths shown in the figure to the right, from the least to the greatest: a) a, b, c, d b) b, d, a, c c) c, d, b, a d) c, b, a, d e) d, c, a, b Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Quick Answer Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed paths shown in the figure to the right, from the least to the greatest: b) b, d, a, c Remember: 𝑩. 𝑑𝒔 = 𝜇0 𝐼 Ampere’s law indicates that the value of the line integral depends only on the net current through each closed path. Path b encloses 1 A, path d 3 A, path a 4 A and path c encloses 6 A Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Quick Quiz Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed paths shown in the figure to the right, from the least to the greatest: a) a, b, c, d b) b, c, d, a c) b, d, a, c d) d, c, a, b e) The criteria are badly chosen, in this case Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Quick Answer Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed paths shown in the figure to the right, from the least to the greatest: e) The criteria are badly chosen, in this case Ranked from least to greatest, first comes b, but then a = c = d. These three paths all give the same non-zero value 𝑩. 𝑑𝒔 = 𝜇0 𝐼 since the size and shape of the paths do not matter. Path b does not enclose the current, and hence its line integral is zero. Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Field Due to a Long Straight Wire – From Ampere’s Law Calculate the magnetic field at a distance r from the centre of a wire that carries a steady current, I. The current is uniformly distributed through the cross section of the wire. Since the wire has a high degree of symmetry, the problem can be categorised as an Ampere’s Law problem. For r ≥ R, the result should be the same as that obtained using the Biot-Savart law. Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Field Due to a Long Straight Wire – From Ampere’s Law Outside the wire, r > R 𝑩. 𝑑𝒔 = 𝐵. 2𝜋𝑟 = 𝜇0 𝐼 ∴ 𝐵= 𝜇0 𝐼 2𝜋𝑟 Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Field Due to a Long Straight Wire – From Ampere’s Law Inside the wire, we need I’, which is the current Inside the amperian circle: 𝐼′ = 𝜋𝑟 2 𝐼 𝜋𝑅 2 Using this: 𝑩. 𝑑𝒔 = 𝐵 2𝜋𝑟 = 𝜇0 𝐼′ Substitute in for I’, giving us: 𝐵 2𝜋𝑟 = 𝑟2 𝜇0 2 𝐼 𝑅 Electricity & Magnetism – Magnetism → 𝐵= 𝜇0 𝐼 2𝜋𝑅2 𝑟 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields Field Due to a Long Straight Wire – Results Summary r<R 𝐵= r>R 𝐵= 𝜇0 𝐼 2𝜋𝑅 2 𝑟 𝜇0 𝐼 2𝜋𝑟 The field is proportional to r, inside the wire. The field varies as 1/r outside the wire. The two equations are equal at r = R. Much easier to calculate using Ampere’s law rather than the Biot-Savart law! Electricity & Magnetism – Magnetism 22/08/2013 PHYSICS 1B – Ampere's Law and Magnetic Fields The Magnetic Field of a Toroid Find the field at a distance r from the centre of the toroid, which has N turns of wire. The wire passes through the loop N times, so the total current through the loop is N times I. 𝑩. 𝑑𝒔 = 𝐵. 2𝜋𝑟 = 𝜇0 𝑁𝐼 ∴ 𝜇0 𝑁𝐼 𝐵= 2𝜋𝑟 Electricity & Magnetism – Magnetism 22/08/2013