Homework VI - PHY2061
1.) In class we considered a problem (B field from two wires in some ‘symmetric geometrical arrangement’) like sample problem 33-3. Consider instead the arrangement below, where two long wires carring currents (both equal to ‘i’) are coming out of the diagram, with the x-y plane in the diagram. What is Bx, the total field in the x-direction, at point P from the B-field contributions from both wires taken together, i. e. Bx=Bx1+Bx2? Please include the sign of Bx. (The two currents both go thru the x-axis, a distance of 3 a apart, while point P is in the y-direction above the right wire a distance of 4 a.)
solution: magnitude Blong wire = µ0i/2d. From the lower left wire, the distance to P is 5 a, and the (negative) x-component of B is µ0i/2(5a) cos36.9o. From the lower right wire, the B field is all in the negative x direction, with magnitude µ0i/2(4a). Thus, the total magnitude in the – x direction is (µ0i/2)(cos36.9o/5 a + 1/4a) =
0.41(µ0i/2a)
2. Suppose there is an ideal solenoid, with 1000 turns/cm (beware units.)
How many amperes of current must flow in this ideal solenoid to create a field of 1 T in the center of the solenoid? B=µ0ni , so 1 T = 4 10-7 1000*100 i, implies that i=7.96 A
3. Using the current from problem 3 above, we want to calculate the Joule heating (=i2R) in the coil from the energy loss of the current flowing through the resistance of the wire. To do this, knowing the resistivity ρ of copper wire is 1.69 10-8 Ωm from chapter 29, you need to find the total resistance R of the wire from R=ρL/A, where A is the cross sectional area of the wire (r2 or d2/4) and L is the total length of the wire.
a. Assume a 10 cm long coil with a 1 cm inner diameter (close to an “ideal” solenoid), i. e. a total of 1000 turns/cm * 10 cm = 10,000 turns of wire. Assume 40 gauge wire, diameter d= 0.08 mm (beware units), so there are 125 (=1 cm/0.08mm) turns in one layer per cm, so to get 1000 turns/cm you have to have 8 layers of wire. If the solenoid on which the wire is wrapped is 1 cm diameter, 8 layers of 0.08 mm diameter wire make the outer diameter of the 8 layers of wire wrapped around the solenoid about 1.064 cm. Assume therefore an average wire coil diameter of 1.032 cm, so *each* turn has a length of 2r or d=3.24 cm (beware units.) On the entire solenoid, there are 10,000 turns.
Calculate the total length L of the wire wrapped around the solenoid.
L=3.24 10-2m * 10,000 = 324 m
b.) calculate the total R of the wire in the solenoid. R=1.69 10-8 324/((8 10-5)2/4)=
0.109 104 Ω
c.) what is the Joule heating? (If you’ve kept track of your powers of 10, you should get units of kilowatts. This is why you don’t generate a 1 T field from copper wire!)
i2R=7.962 * 1090 = 69.1 kW
4. In the text, and in class, there was a discussion of the magnetic field inside a wire using Ampere’s law with the assumption that “a uniform distribution” of the current in the wire gives the fraction of the current inside a radius ‘r’, where the diameter of the wire is R and r<R, as being r2/R2. Suppose instead that the current is distributed in a wire such that the current inside a radius ‘r’ is proportional to r/R. Calculate the field inside the wire as a function of ‘r’ and draw a graph of your result. Ampere’s law (see also p. 762) gives B around a circle in the wire * 2r (radius of circle) = µ0i (r/R), so the distance from the center of the wire ‘r’ cancels and the field is constant inside the wire with this ‘uniform distribution.’