Magnetic field for a wire with non-uniform current density (Ampere's Law example).
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In this Ampere's Law example,
In this Ampere's Law example, we compute the magnetic field for a wire with non-uniform current density. We are given a linearly increasing current density J=kr, and we want to express the magnetic field inside and outside the wire in terms of the total current I.
We start with a picture of the wire in perspective with an Amperian loop shown inside the wire and the statement of Ampere's Law, then we get a quick reminder of the definition of current density and how to find current from current density. We quickly realize that we can't compute the enclosed current within the Amperian loop by simple multiplication, because the current density doesn't have a single well defined value over the enclosed surface.
So our strategy is to rewrite the formula I=JA in infinitesimal form as dI_enc=J*dA, where the dA's are the areas of thin rings centered on the symmetry axis of the wire. Over the infinitesimal area of each thin ring, the value of the current density J is now well-defined as kr', where we use an r' to distinguish from the radius of the Amperian loop, r. Now each thin ring has an infinitesimal area 2pi*r'*dr', and we can complete the setup of our expression for dI_enc.
Next, we sum the current contributions by using an integral (remember, integrals are really just continuous summation devices!). We compute the integral and find the enclosed current in terms of the constant k, but we still want to find the magnetic field inside the wire in terms of the total current I, not k!
The trick to getting our enclosed current in terms of I is to do a second integral of current density over the entire cross sectional area of the wire. The result of that integral is I, and when we perform the integral we get an expression for I in terms of k, allowing us to substitute for k in terms of I in the expression for I_enc.
Now we have the right hand side of Ampere's law, and we can get back to finding the magnetic field: the path integral on the left hand side of Ampere's law is trivial because the magnetic field points in the same direction as the path increments dl, and the magnetic field magnitude is constant along the Amperian loop as well. This means the path integral reduces to B(2pi*r), and we're ready to solve for B. Setting B(2pi*r) equal to mu_0 times the enclosed current, we can cancel a factor of r, and we find that the magnetic field is increasing quadratically within the wire.
For completeness, we also find the magnetic field outside the wire, but this is a trivial calculation because an amperian loop outside the wire contains ALL the current I. So we get the classic result B=mu_0*I/2pi*r outside the wire.
Finally, we show that the magnetic field is continuous at the surface of the wire, and we produce a plot of magnetic field in and out of the wire.
We start with a picture of the wire in perspective with an Amperian loop shown inside the wire and the statement of Ampere's Law, then we get a quick reminder of the definition of current density and how to find current from current density. We quickly realize that we can't compute the enclosed current within the Amperian loop by simple multiplication, because the current density doesn't have a single well defined value over the enclosed surface.
So our strategy is to rewrite the formula I=JA in infinitesimal form as dI_enc=J*dA, where the dA's are the areas of thin rings centered on the symmetry axis of the wire. Over the infinitesimal area of each thin ring, the value of the current density J is now well-defined as kr', where we use an r' to distinguish from the radius of the Amperian loop, r. Now each thin ring has an infinitesimal area 2pi*r'*dr', and we can complete the setup of our expression for dI_enc.
Next, we sum the current contributions by using an integral (remember, integrals are really just continuous summation devices!). We compute the integral and find the enclosed current in terms of the constant k, but we still want to find the magnetic field inside the wire in terms of the total current I, not k!
The trick to getting our enclosed current in terms of I is to do a second integral of current density over the entire cross sectional area of the wire. The result of that integral is I, and when we perform the integral we get an expression for I in terms of k, allowing us to substitute for k in terms of I in the expression for I_enc.
Now we have the right hand side of Ampere's law, and we can get back to finding the magnetic field: the path integral on the left hand side of Ampere's law is trivial because the magnetic field points in the same direction as the path increments dl, and the magnetic field magnitude is constant along the Amperian loop as well. This means the path integral reduces to B(2pi*r), and we're ready to solve for B. Setting B(2pi*r) equal to mu_0 times the enclosed current, we can cancel a factor of r, and we find that the magnetic field is increasing quadratically within the wire.
For completeness, we also find the magnetic field outside the wire, but this is a trivial calculation because an amperian loop outside the wire contains ALL the current I. So we get the classic result B=mu_0*I/2pi*r outside the wire.
Finally, we show that the magnetic field is continuous at the surface of the wire, and we produce a plot of magnetic field in and out of the wire.
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در تاریخ 1403/03/31 منتشر شده
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