Question

Problem 1) Solve parts d and e Also given: Magnitude of force on the wire Fb 90N, in B-3.0T 1. A L 5.0 m piece of wire carrying a current I-6.0 A runs along the -j direction. It has an x coordinate of 0 m, and is at a height of z 1.0 m above the origin. 2-d view 3-d view +z ty TX +x (d) (2 pts) The center of the wire is tethered to the origin with a 1.0 m piece of non- conducting string (F-1 m k. Calculate the torque about the origin (magnitude and direction) due to the magnetic force on the wire. (e) (2 pts) The wire is turned so that it now points along i. What is the magnitude and direcetion of the resulting magnetic force on the wire?

Magnitude and Direction will have numerical values. Use equations to solve.

Answer 1

d. Magnetic force acting on straight wire is along +Z direction so its line of action is passing through the origin hence perpendicular distance of line of action of this force from origin is zero. Hence torque = Force × perpendicular distance of line of action of the force from origin = 0

e. As magnetic field is already along X axis and now current carrying wire is also along X axis hence angle between them is 0⁰ hence force = ILB sin 0⁰ = 0