A uniform straight metal wire of length 2.0 m is released from rest and falls horizontally under gravity ( g = 10 ~m/s^2 ) in a region with a uniform horizontal magnetic field of 0.4 Gauss, which is perpendicular to the length of the wire. The time elapsed from the release of the wire when the induced EMF across it reaches 1.6 mV is:

Step-by-step solution

The induced motional EMF in the falling wire is given by E = BLv , where v is the instantaneous velocity. Given: Induced EMF, E = 1.6 mV = 1.6 10⁻³ V Magnetic field, B = 0.4 Gauss = 0.4 10⁻⁴ T = 4 10⁻⁵ T Length of wire, L = 2.0 m Substituting the values to find velocity v : 1.6 10⁻³ = (4 10⁻⁵) 2.0 v 1.6 10⁻³ = 8 10⁻⁵ v v = 1.6 10⁻³ 8 10⁻⁵ = 160 10⁻⁵ 8 10⁻⁵ = 20 m/s For a body falling freely from rest, the velocity after time t is v = gt . 20 = 10 t t = 2 s Answer: 2 s