Question

A long vertical wire carries a steady 211 A current. A pair of horizontal rails are 0.20 m apart. A 20-1 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 horizontal velocity of 0.90 m/s to the right, as shown. At the instant that the bar is 0.20 m from the wire, what are the induced current in the resistor and its direction through the resistor? 2.00 m 0.20 m a v = 0.90 m/s 211 A 0.20 m 2012 * b A. B. C. 6.3 x 10-7 A from a to b 6.3 x 10-7 A from b to a 3.2 x 10-7 A from a to b 3.2 x 10-7 A from b to a 1.9 x 10-6 A from a to b 1.9 x 10-7 A from b to a D. E. F.

Answer 1

At the instant the bar is 0.20 m from the wire, an induced current is created in the resistor due to the changing magnetic field caused by the moving bar. To determine the induced current and its direction through the resistor, we can use the formula for induced current: Induced current = (magnetic field strength) x (length of conductor) x (velocity of conductor) The magnetic field strength at a distance r from a long vertical wire carrying current I is given by: Magnetic field strength = (μ₀ * I) / (2πr) where μ₀ is the permeability of free space. Plugging in the given values: Current I = 211 A Distance r = 0.20 m Substituting these values into the formula for magnetic field strength, we get: Magnetic field strength = (4π * 10^-7 T m/A * I) / (2π * r) Simplifying further: Magnetic field strength = (2 * 10^-7 T m/A * I) / r Now, we need to determine the direction of the induced current through the resistor. According to Lenz's law, the direction of the induced current will be such that it opposes the change that caused it. Since the bar is moving to the right, the induced current will flow in the opposite direction, from point b to point a. Substituting the given values: Length of conductor = 0.20 m Velocity of conductor = 0.90 m/s Substituting these values into the formula for induced current, we get: Induced current = (2 * 10^-7 T m/A * 211 A * 0.20 m) / 0.90 m/s Simplifying further: Induced current = 4.69 x 10^-6 A Therefore, the main answer is: The induced current in the resistor is 4.69 x 10^-6 A from point b to point a. In conclusion, when the bar is 0.20 m from the wire, there is an induced current of 4.69 x 10^-6 A flowing from point b to point a through the resistor.