Question

A student is measuring the magnetic field generated by a long, straight wire carrying a constant current. A magnetic field probe is held at various distances d from the wire, as shown above, and the magnetic field is measured. The graph below shows the five data points the student measured and a best-fit curve for the data. Unfortunately, the student forgot about Earth’s magnetic field, which has a value of 5.0 x 10^-5 T at this location and is directed north.

(a) On the graph, plot new points for the field due only to the wire.

(b) Calculate the value of the current in the wire.

Another student, who does not have a magnetic field probe, uses a compass and the known value of Earth’s magnetic field to determine the magnetic field generated by the wire. With the current turned off, the student places the compass 0.040 m from the wire, and the compass points directly toward the wire as shown below. The student then turns on a 35 A current directed into the page.

(c) On the compass, sketch the general direction the needle points after the current is established.

(d) Calculate how many degrees the compass needle rotates from its initial position pointing directly north.

The wire is part of a circuit containing a power source with an emf of 120 V and negligible internal resistance.

(e) Calculate the total resistance of the circuit.

(f) Calculate the rate at which energy is dissipated in the circuit.

Answer 1

(a) Based on the RHR for a current wire creating a field, the magnetic field at the location in question is directed down (south). So the students reading was less than it should have been for the wire since the meter was measuring both fields and the earth field was acting against the wires field. If the earth’s field was not present, the meter reading would have been larger. So each Field data point should be shifted up by the amount that the earth field had reduced it reading, as shown.

(b) The field in the wire is given by B = μ_oI / (2πd). Use one of the new data points. 10x10^–4 = 4π x 10^–7 (I) / 2 π(0.01) I = 50 A 

(c) To figure out the direction the needle points, we first need to determine how strong the wires field is with the 35 A current in the given location. B = μ_oI / (2πd) … B = 4π x 10^–7 (35) / 2 π (0.04) … B = 17.5 x 10^–5 T, as compared to the earth field given in the problem, 5x10^–5 T. Since the fields are on the same order of magnitude, the compass will not be totally overpowered by the external field and will point in a North–westerly direction though it would be angled more towards west since the external western field is larger.

(d) Using the value of the earth field, and the value of the external field, we can determine the exact angle by setting them up head to tail.

The angle θ is found using tan θ = o/a

tan θ = 17.5x10^–5 / 5x10^–5

θ = 74 degrees.

(e) V = IR … 120 = 35 R … R = 3.4 Ω

(f) Rate of energy is power. P = IV = (35)(120) = 4200 W