Question
A long vertical wire carries a steady 6 A current. A pair of rails are horizontal and are 0.3 m apart. A 11.6 O resistor connects points a and b, at the end of the rails. A bar is in contact with the rails, and is moved by an external force with a constant velocity of 0.4 m/s as shown. The bar and the rails have negligible resistance. At a given instant t1, the bar is 0.28 m from the wire, as shown. In Figure, at time t1, the induced current in µA is:
Express the answer with three decimal places.
Explanation / Answer
wire carrying current I = 6A distance between the pairs of Horizontial rails, d = 0.3 m Moving velocity of the bar, V = 0.4 m/s At given instant of time bar at a distance, d = 0.3 m from the rails Let ,induced current = Iinduced Concept: when current passing through the wire an amount of magnetic field will induced around it . It is known by the formula , for magnetic filed generated around the wire is B = o I / 2 d =( 4 x10-7 )(6 A)/2 (0.3) = 40 x10-7 T We have a formula for emf induced in the wire while it moving with velocity in the magnetic field e.m.f or voltage = BdV = (40 x10-7 T)(0.3)(0.4) = 4.8 x10-7 V but , resistence of the wire = R = 11.6 thus, amount of current induced in the wire would be = Iinduced = (4.8 x10-7 V) / 11.6 = 4.14 x10-8 A thus, current induced in the wire is Iinduced = 0.0414 A when current passing through the wire an amount of magnetic field will induced around it . It is known by the formula , for magnetic filed generated around the wire is B = o I / 2 d =( 4 x10-7 )(6 A)/2 (0.3) = 40 x10-7 T We have a formula for emf induced in the wire while it moving with velocity in the magnetic field e.m.f or voltage = BdV = (40 x10-7 T)(0.3)(0.4) = 4.8 x10-7 V but , resistence of the wire = R = 11.6 thus, amount of current induced in the wire would be = Iinduced = (4.8 x10-7 V) / 11.6 = 4.14 x10-8 A thus, current induced in the wire is Iinduced = 0.0414 A = 4.14 x10-8 A thus, current induced in the wire is Iinduced = 0.0414 A