An east-west oriented conducting bar 2.0 m long moves north at 20 m/s through a magnetic field of 0.017 T, directed downward. What is the strength and direction of the electric field generated in the bar?
To determine the strength and direction of the electric field generated in the bar, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the electromotive force (EMF) induced in a conductor is equal to the negative rate of change of the magnetic flux within the circuit.
The magnetic flux, denoted by Φ, is the product of the magnetic field strength (B) and the area (A) through which the magnetic field passes. In this case, since the bar is moving perpendicular to the magnetic field, the area through which the magnetic field passes is given by the length of the bar (2.0 m) multiplied by its width (the width is not given, so let's assume it to be 1 meter for simplicity).
Therefore, the magnetic flux (Φ) can be calculated as:
Φ = B * A = B * l * w,
where B = 0.017 T (given), l = 2.0 m (length of the bar), and w = 1 m (assumed width).
Next, we need to calculate the rate of change of magnetic flux (∆Φ/∆t) since the bar is moving. The rate of change of magnetic flux is equal to the product of the velocity (v) of the bar and the width (w) of the bar:
∆Φ/∆t = v * w = 20 m/s * 1 m = 20 m²/s.
Now, we can calculate the EMF induced in the bar using Faraday's law:
EMF = -∆Φ/∆t,
EMF = -(20 m²/s).
Finally, since the bar is a conductor, the induced EMF creates an electric field within the bar. The strength and direction of the electric field can be determined by dividing the EMF by the length of the bar (l):
Electric field strength (E) = EMF / l,
E = -(20 m²/s) / 2.0 m = -10 V/m.
Therefore, the strength of the electric field generated in the bar is 10 V/m, and it is directed from east to west (-x direction) within the bar.