two straight and parallel wires are separated by 6 cm. there is a 2 A current flowing in the first wire. if the magnetic field strength is found to be zero between the two wires at a distance of 2.2 cm from the first wire, what is the magnitude and direction of the current in the second wire?
If the magnetic field at 2.2cm is zero it means that at that distance the fields produced by the two wires cancel out. At 2.2cm from the first wire the field strength is B= uI/ 2pi r ... . where u is the permitivity of free space, 4.0 x 10^-7 N/A^2 B= (4.0 x 10^-7 N/A^2)( 2 A)/ 2(3.14)(0.022m) B=( 8 x10^-7)/(0.138) B= 5.79 x 10^-6 Tesla The magnitude of the magnetic field from the other wire must be equal in magnitude at 3.8cm away (6- 2.2) and opposite in direction. Using the right hand rule (thumb points in direction of current and fingers curl around the wire to show the direction of the magnetic field) you can see that for the magnetic fields to be in opposite direction between the two wires the current must be flowing in the same direction as the first wire. Using the formula for magnetic field strength you can then determine the current needed. 5.79 x10^-6 T= (4.0 x 10^-7) I/ 2 (3.14)(0.038m) THen solve for I I= (5.79 x 10^-6 T)(2)(3.14)(0.038m)/ 4.0 x 10^-7) I= 3.45 A Therefore there is a current of 3.45A in the wire and it is flowing in the same direction as the first.
