Using Ampere's Law to find the magnetic field inside and outside a wire.
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To find the magnetic field
To find the magnetic field of a thick wire (magnetic field of a cylindrical conductor), we apply Ampere's Law to find the magnetic field inside a wire and the magnetic field outside a wire by using separate Amperian loops. We're told to assume the wire has a uniform current density (and in future videos we'll relax this assumption).
We make a couple notes on the current density. First, the definition of current density is J=I/A, the current per unit cross sectional area. Second, we can turn this around to find the current as I=J*A, where A is the cross sectional area exposed to the current. Finally, we can calculate the current density in our wire by taking the total current I and dividing by the cross sectional area, which is given by pi(R^2).
Using Ampere's Law to find the magnetic field inside the wire, we start by using the right hand rule to get a sense for the orientation of the magnetic field, which swirls around the symmetry axis of the thick wire. This tells us how to choose an Amperian loop for the path integral in Ampere's law: it should just be a circle sharing the symmetry axis of the cylindrical wire!
When we apply ampere's law, we evaluate the path integral on the left hand side, and in this path integral, the dot product is trivial because the magnetic field vector is always parallel to the path increment vector. In addition, the magnetic field magnitude is constant along the path (by symmetry), allowing us to factor B out of the Ampere's Law path integral. So the Ampere's law integral reduces to B*2pi*r. On the right hand side, we have mu_0*Ienclosed, but Ienclosed is less than the total current in the wire. We have to use the definition of current density and go back to the expression I=J*A, where J is the current density in the cylindrical wire and A is the cross section of the Amperian loop. Plugging these values in and solving for B, we arrive at an expression for the magnetic field inside a cylindrical wire: B=(mu_0*I*r)/(2*pi*R^2), where r is the distance from the axis of the thick wire and R is the radius of the thick wire.
Now we use Ampere's law to compute the magnetic field outside the wire. The path integral is trivial in exactly the same way as the first, quickly reducing to B(2pi*r). On the right hand side of Ampere's Law, we simply get mu_0*I, because the entire current is enclosed within the Amperian loop! Solving for B, we get B=mu_0*I/2pi*r, which is the same as the result for the magnetic field around a thin current carrying wire.
Finally, we want to graph the magnetic field inside and outside a wire as a function of r, but first we want to show that B is continuous at the surface of the cylindrical conductor. We evaluate B_in(R) and B_out(R) and show that they reduce to the same expression, then we show a plot of the magnetic field as a function of distance from the axis of the thick wire. As expected, the magnetic field shows linear growth within the material, then drops off like 1/r outside the wire.
In the next video, we investigate the magnetic field of a thick wire with non-uniform current density, and that requires an extra non-trivial integral in the solution!
We make a couple notes on the current density. First, the definition of current density is J=I/A, the current per unit cross sectional area. Second, we can turn this around to find the current as I=J*A, where A is the cross sectional area exposed to the current. Finally, we can calculate the current density in our wire by taking the total current I and dividing by the cross sectional area, which is given by pi(R^2).
Using Ampere's Law to find the magnetic field inside the wire, we start by using the right hand rule to get a sense for the orientation of the magnetic field, which swirls around the symmetry axis of the thick wire. This tells us how to choose an Amperian loop for the path integral in Ampere's law: it should just be a circle sharing the symmetry axis of the cylindrical wire!
When we apply ampere's law, we evaluate the path integral on the left hand side, and in this path integral, the dot product is trivial because the magnetic field vector is always parallel to the path increment vector. In addition, the magnetic field magnitude is constant along the path (by symmetry), allowing us to factor B out of the Ampere's Law path integral. So the Ampere's law integral reduces to B*2pi*r. On the right hand side, we have mu_0*Ienclosed, but Ienclosed is less than the total current in the wire. We have to use the definition of current density and go back to the expression I=J*A, where J is the current density in the cylindrical wire and A is the cross section of the Amperian loop. Plugging these values in and solving for B, we arrive at an expression for the magnetic field inside a cylindrical wire: B=(mu_0*I*r)/(2*pi*R^2), where r is the distance from the axis of the thick wire and R is the radius of the thick wire.
Now we use Ampere's law to compute the magnetic field outside the wire. The path integral is trivial in exactly the same way as the first, quickly reducing to B(2pi*r). On the right hand side of Ampere's Law, we simply get mu_0*I, because the entire current is enclosed within the Amperian loop! Solving for B, we get B=mu_0*I/2pi*r, which is the same as the result for the magnetic field around a thin current carrying wire.
Finally, we want to graph the magnetic field inside and outside a wire as a function of r, but first we want to show that B is continuous at the surface of the cylindrical conductor. We evaluate B_in(R) and B_out(R) and show that they reduce to the same expression, then we show a plot of the magnetic field as a function of distance from the axis of the thick wire. As expected, the magnetic field shows linear growth within the material, then drops off like 1/r outside the wire.
In the next video, we investigate the magnetic field of a thick wire with non-uniform current density, and that requires an extra non-trivial integral in the solution!
4 هفته پیش
در تاریخ 1403/03/29 منتشر شده
است.
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