Question

A 2.00-m length of wire is held in an east–west direction and moves horizontally to the north with a speed of 0.500 m/s. The Earth’s magnetic field in this region is of magnitude 50.0 mT and is directed northward and 53.08 below the horizontal. 

(a) Calculate the magnitude of the induced emf between the ends of the wire and (b) determine which end is positive.

Answer 1

(a) The motional emf induced in a conductor is proportional to the component of the magnetic field perpendicular to the conductor.

\(\varepsilon=B_{\perp} l v\)

\(=\left[\left(50.0 \times 10^{-6} \mathrm{~T}\right) \sin 53.08^{\circ}\right](2.00 \mathrm{~m})(0.500 \mathrm{~m} / \mathrm{s})\)

\(=3.99 \times 10^{-5} \mathrm{~V}\)

Thus, the magnitude of the induced emf between the ends of the wire is \(39.9 \mu \mathrm{V}\).

(b) Use the right-hand rule as follows:

The magnetic force on charges in the wire would tend to move positive charges westward.

The West endis positive