Question

In Oersted's experiment, suppose that the compass was 0.15 m from the current-carrying wire.

If a magnetic field of one third the Earth's magnetic field of  5.0×10−5T was required to give a noticeable deflection of the compass needle, what current must the wire have carried?

Express your answer using two significant figures.

Answer 1

Firstly, the Tesla (T) is a unit of magnetic flux density (B), not of magnetic field strength (H). The unit of field strength is Amps/m (A/m). It is important in all scientific studies to be absolutely clear about the distinctions between the various quantities being discussed, their units, and symbols.

Secondly, we are not told enough about the disposition of the current-carrying wire to be able to estimate properly the flux density in the space around it. If, for the sake of simplicity we take the impractical case of a wire which is is straight and infinite in extent we can use the formula -

B = u*I/(2*pi*r)

Where u is the permeability of air (taken to be the same as that of vacuum) r is the distance from the wire, I is the current. Clearly this will only relate approximately to a practical experiment of the kind that Oersted did. An alternative (more practical) assumption would be that the wire is in the form of a circular loop. But then we would need to know a lot more extra information, including the size of the loop, the exact position of the compass relative to the loop and the orientation of the plane of the loop relative to the earth's magnetic field direction.

A further point is that B is a vector quantity, and its effect on a compass needle depends on the direction of the component of B created by the wire in relation to the component of B due to the earth's magnetism. If these two components happen to be in the same direction, the needle would not be deflected at all however large the current might be. To pretend that a value of B of 1.66T is

'.

Using B = u*I/(2*pi*r) ....

I = 2*pi*r*B/u = 2*pi*0.15*1.66*10^-5/1.25*10^-6 = 12.5161A