One long wire carries current 27.0 A to the left along the x axis. A second long wire carries current 71.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of -2.00 JC is moving with a velocity of 1501 Mm/s along the line. Ignore relativistic effects. (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field. Part 1 of 8 - Conceptualize We know that wires carrying current cause deflection of moving charged particles in a vacuum tube. We estimate that the magnetic fields of the two currents in this problem may add to zero at a location some tens of centimeters below the first wire. The force may be a small fraction of a newton and it is important to determine its direction. The strength of the electric field may be on the order of kilonewtons per coulomb. Part 2 of 8 - Categorize We must find a location, farther away from the wire carrying the larger current and closer to the wire carrying the smaller current, where the fields created separately by the currents are equal in magnitude but opposite in direction. We will compute the magnitude and identify the direction of the magnetic field each current creates at the location between them. We will add the fields as vectors and find the magnetic force on the moving charge from a cross product. In the last part of the problem, we will review the definition of the electric field by finding an electric field whose force counterbalances the force on the moving charge so that the charge moves undeflected through the fields. Part 3 of 8 - Analyze (a) Above the pair of wires, the field out of the page of the 12 = 71.0 A current will be larger than the field into the page of the I, = 27.0 A current, so they cannot add to zero. The two wires separated by distance a are labeled I, and I2, with the x, y, and z axes shown in the diagram below. Between the wires, both produce fields into the page. Their magnetic fields can only add to zero below the wires, at coordinate y = -lyl. Here, the total field is given by the following. B = HO (2 ) (-*) + 2 (1)] = 0 . z A Simplifying, we have 12lyl= 1, (ly|+ a). Solving for the distance below I, remembering that this wire is at y = 0 so this coordinate is negative, we have Y = 1 1 - 12m) A) -C The net field is zero all along the line at y = - m