A 3-cm length of wire is moved at right angles across a uniform magnetic field with a speed of 2.0 m/s. If the flux density is 5.0 teslas, what is the magnitude of the induced e.m.f?

Explanation

To calculate the induced electromotive force (EMF) in this scenario, we can use Faraday's Law of Electromagnetic Induction, which states that the induced EMF is equal to the negative rate of change of magnetic flux. Formula:EMF = - (dΦ/dt)Where: EMF is the induced electromotive force (in volts) dΦ/dt is the rate of change of magnetic flux (in webers per second)Breaking down the problem:Magnetic flux (Φ): The magnetic flux is given by the product of the magnetic field strength (B) and the area (A) perpendicular to the field. In this case, the wire is moving perpendicular to the field, so the area is simply the length of the wire times its width (which we'll assume is negligible compared to its length). Φ = B * A = B * L * wRate of change of magnetic flux: Since the wire is moving at a constant speed, the area it sweeps through per second is constant. Therefore, the rate of change of magnetic flux is simply the product of the magnetic field strength and the speed of the wire. dΦ/dt = B * L * vSubstituting values:B = 5.0 TL = 3 cm = 0.03 mv = 2.0 m/sCalculation: EMF = - (dΦ/dt) = - (B * L * v) = - (5.0 T * 0.03 m * 2.0 m/s) = -0.3 VThe magnitude of the induced EMF is 0.3 volts.Therefore, the correct answer is Option 2: 0.30 V.