Review A charged particle with charge q and mass m moving with speed v perpendicular to a constant magnetic field with magnitude B will travel in a circle with radius r given by Physics for Scientists Engineers 2 r mv qB For the same conditions we can relate the momentum p and the charge q to the magnitude of the magnetic field B and the radius r of the circular motion Spring Semester 2005 Lecture 22 Br February 24 2005 Physics for Scientists Engineers 2 1 February 24 2005 Review 2 2 Let s address the problem of calculating magnetic fields generated by moving charges Remember that we calculated the electric field in terms of the electric charge using the form dE V V dhne VH hne B H H dv di i 1 dq 4 0 r 2 where dq is a charge element where n is the number of electrons per unit volume and e is the charge of an electron The electric field points radially from the electric charge so that we can write Hall Effect dE Physics for Scientists Engineers 2 Physics for Scientists Engineers 2 Magnetic Fields If we run a current i through a conductor of width h in a constant magnetic field B we induce a voltage VH across the conductor that is given by February 24 2005 p q 3 February 24 2005 1 dq r 4 0 r 3 Physics for Scientists Engineers 2 4 1 Magnetic Fields 2 Magnetic Fields 3 The direction of the magnetic field produced by the current element is perpendicular to both the radial direction and to the current element The magnetic field has an added complication because the current that produces the magnetic field has a direction The electric field is produced by charge that is a scalar The magnitude of the magnetic field is given by We can write the magnetic field produced by a current element ids as ids r dB 0 4 r 3 dB This formula is the Biot Savart Law 0 is the magnetic permeability of free space whose value is Tm 0 4 10 7 A February 24 2005 Physics for Scientists Engineers 2 5 Magnetic Field from a Long Straight Wire 0 ids sin 4 r 2 Where is the angle between the radial direction and the current element Let s calculate the magnetic field for various current element distributions February 24 2005 Physics for Scientists Engineers 2 6 Magnetic Field from a Long Straight Wire 2 For our first application we will calculate the magnetic field from an infinitely long straight wire We calculate the magnetic field dB at a point P at a distance r from the wire as illustrated below We will calculate the magnetic field from the right half of the wire and multiply by two to get the magnetic field from the whole wire The magnitude of the magnetic field from the right side of the wire is given by ids sin i ds sin B 2 dB 2 0 0 2 0 0 4 r 2 0 r 2 We can relate r s and by The magnitude of the magnetic field will be given by the Biot Savart Law and the direction will be out of the page February 24 2005 Physics for Scientists Engineers 2 r s 2 r 2 7 February 24 2005 sin r s r 2 2 Physics for Scientists Engineers 2 8 2 Magnetic Field from a Long Straight Wire 3 Magnetic Field from a Long Straight Wire 4 Substituting we get B The direction of the magnetic field is given by the right hand rule 0i r ds 2 0 s 2 r 2 3 2 If you grab the wire such that your thumb points in the direction of the current your fingers will point in the direction of the magnetic field as shown Carrying out this integral we get i 1 r s i 0 B 0 2 1 2 2 r s 2 r 2 2 r 0 s s 2 r 2 1 2 10 s0 Looking down the wire the magnetic field lines form circles So our resulting equation for the magnetic field at a perpendicular distance from a long straight wire carrying a current is B r 0i 2 r February 24 2005 Physics for Scientists Engineers 2 9 Magnetic Field from a Loop 10 Going around the loop we can relate an angle to the current element by dS Rd allowing us to calculate the magnetic field at the center of the loop We start with B dB ids sin dB 0 4 r 2 2 0 0 iRd 0i 4 R 2 2R Please keep in mind that this calculation only gives us information on the value of the magnetic field at the center of the loop and apply it to this case We can see that r R and 90 for every current element along the loop Using other techniques we can calculate the magnetic field everywhere is shown to the right For the magnetic field from each current element we get ids sin 90 0 ids dB 0 4 R2 4 R 2 Physics for Scientists Engineers 2 Physics for Scientists Engineers 2 Magnetic Field from a Loop 2 Let s calculate the magnetic field at the center of a circular loop of wire carrying current February 24 2005 February 24 2005 11 February 24 2005 Physics for Scientists Engineers 2 12 3 Ampere Ampere s Law Ampere Ampere s Law 2 Recall that we can calculate the electric field resulting from any distribution of electric charge using 1 dq dE r 4 0 r 3 In a similar way we can calculate the magnetic field from an arbitrary distribution of current elements using If the charge distribution were complicated we would be faced with a difficult integral In cases where the distribution of current elements has cylindrical or spherical symmetry we can apply Ampere s Law to calculate the magnetic field from a distribution of current elements with much less effort than using a direct integration ids r dB 0 4 r 3 However we again may be faced with a difficult integral However if the charge distribution had cylindrical or spherical symmetry we could apply Gauss Law 0 E dA q Ampere s Law is 0 enc Where the integral is carried out around an Amperian loop and ienc is the current enclosed by the loop and obtain the electric field in an elegant manner February 24 2005 Physics for Scientists Engineers 2 B ds i 13 Ampere Ampere s Law 3 February 24 2005 Physics for Scientists Engineers 2 14 Magnetic Field inside Long Wire As an example of Ampere s Law consider the five currents shown below Consider a current i flowing out of the page in a wire with a circular cross section of radius R This current is uniformly distributed over the cross sectional area of the wire The currents are perpendicular to the plane shown To find the magnetic field we use an Amperian loop with radius represented by the red circle We can draw an Amperian loop …
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