1. A semi circular wire of radius b is carrying a current I in the clockwise dir
ID: 2260110 • Letter: 1
Question
1. A semi circular wire of radius b is carrying a current I in the clockwise direction as shown below. The center of the semi circle is located at (0,b). The goal of this problem is to apply the Biot-Savart law to set up the integral for the magnetic field at the origin. This problem is designed to show you haw to apply the Biot-Savart law. Please complete parts a thru g in order.
a.
Using the law of cosines, determine the expression for |r|2, the magnitude of the vector that is directed from the current element to the origin.
b. Write the expression that relates the angle ? to the angle ?.
c. Write the expression for the unit vector, [r HAT] , that is directed from the current element to the point where the magnetic field is determined. Express the unit vector in terms of ?.
d. Write the expression for the vector [phi]. Express this vector in terms of ?.
e. Determine ds x r [ds cross r HAT].
f. Now that we have all elements of the integral, write the expression for the integral.
g. Use an online resource to evaluate the definite integral. Express your result in terms of I, ?o, ?, b, and the numerical value of the integral in part f.
Explanation / Answer
a)
cos(phi) = (r/2)/b
==> r = 2 b cos(phi)
b)
phi + phi + theta + 90 = 180
==> phi = 45 - theta/2
c)
r^ = sin(phi) i^ - cos(phi) j^
d)
phi^ = cos(phi) i^ - sin(phi) j^ = cos(45 - theta/2) i^ - sin(45 - theta/2) j^
e)
ds x r = |ds| |r| sin(90+phi) = |ds| (2 b cos(phi)) sin(90 + phi)
we know: sin(90+phi) = cos(phi)
==> ds x r = |ds| (2 b cos(phi)) cos(phi)
==> ds x r = 2 b (cos(phi))^2 |ds|
==> ds x r = 2 b (cos(45 - theta/2))^2 |ds|
|ds| = b d(theta)
==> ds x r = 2 b (cos(45 - theta/2))^2 (b d(theta))
==> ds x r = 2 b^2 (cos(45 - theta/2))^2 d(theta)
f)
B = u0/4pi int{i (ds x r)/|r|^3}
==> B = u0/4pi int{i (2 b^2 (cos(45 - theta/2))^2 d(theta))/(2 b cos(45-theta/2))^3}
==> B = u0 i b/4pi int{d(theta)/(cos(45-theta/2))}
g)
B = u0 i b/4pi int{d(theta)/(cos(45-theta/2))} ; integration from 0 to pi
==> B = u0 i b/4pi (3.55)
==> B = 0.881 u0 i b/pi
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