The net force on a current loop in a uniform magnetic field is zero. But what if B is not uniform? The figure (Figure 1) shows a square loop of wire that lies in the xy-plane. The loop has corners at (0, 0), (0, L), (L, 0), and (L, L) and carries a constant current I in the clockwise direction. The magnetic field has no x-component but has both y- and z- components: B = (Boz/L) + (Boy/L) k, where Bo is a positive constant. Figure y (0, L) (0, 0) ½ (L, L) (L, 0) 1 of 1 X Part F Find the direction of the force on side 1, that runs from (L, L) to (L, 0). O +x-direction --direction O+y-direction -y-direction O+z-direction. 0 -z-direction 0 Submit Previous Answers ✓ Correct Part G Find the magnitude of the force on side 4, that runs from (L, 0) to (0, 0). Express your answer in terms of the variables Bo, L, and I. 195 ΑΣΦ ?

Part A through D Please, and thank you A -4.60μC charge is moving at a constant speed of 6.80×10^5m/s in the +x−direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points. Find Bx, By, Bz for each part. Part A x=0.500m, y=0, z=0 Part B x=0, y=0.500m, z=0 Part C x=0.500m, y=0.500m, z=0 Part D x=0, y=0, z=0.500m

An infinitely large slab of thickness d lies on the xy-plane between d, as shown in the figure. (The slab is extended to ±x, and ±y-infinity.) A non-uniform current flows through the slab O and z = with S Jo G) ê 0 d.

A beam of protons is directed in a straight line along the +z direction through a region of space in which there are crossed electric and magnetic fields. The electric field is 450 V/m in the -y direction and the protons move at a constant speed of 10^5 m/s. 1)What must be the magnitude of the magnetic field such that the beam of protons continues along its straight-line trajectory? (Express your answer using two significant figures.)

Problem 4. A Perfect Electric Conductor (PEC) will not support time-varying Electromagnetic Fields (see diagram at left). H₁ = 0 PEC Z H2 = *Hx a.) For this given geometry and using H₁ = 0, with the boundary conditions for magnetic fields on a PEC, determine the value (in terms of Hx) and direction of the surface current, surf, on the PEC. b.) What are the units of Hy and Jsurf?

A particle has a charge of 5 mC and is moving 1000 m/s in the positive x direction. At apoint in time, let the charge be located at the origin. Draw each situation, and draw theposition vector. Find the magnitude and direction of the magnetic field at a pointa. Directly in front of the particle on the x axisb. Directly above the particle a distance 1 cm away (0, 1cm, 0)c. Directly beside the particle at (0, 0, 1cm)d. At coordinates (5 cm, 1 cm, 0)e. At coordinates (5 cm, 0, 1 cm)

An electron with velocity components perpendicular to the y axis is injected into a region of space where the magnetic field is directed towards the +z direction. In addition, the magnetic field lines are straight but has an increasing density towards the +y direction but is not changing in the z or x direction. Describe trajectory and draw a rough illustration of how the particle might behave in this region of space.

A metal cube with sides of length a is moving at velocity v=voj across a uniform magnetic field B = Bok. The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vector of two faces are parallel to the direction of motion). (Figure 1) Figure Bo N 1 of 1 Vo Part A Find E, the electric field inside the cube. Express the electric field in terms of vo, Bo, and unit vectors (i, j, and/or k). E = Submit Part B V ΑΣΦ -î Request Answer Now, instead of electrons, suppose that the free charges have positive charge q. Examples include "holes" in semiconductors and positive ions in liquids, each of which act as "conductors" for their free charges. www. J? If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric field point? Submit Request Answer

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 1.8-cm-wide region of uniform magnetic field in the figure(Figure 1). Figure 1 of 1 d OV 10 kV

Consider the circuit in the figure (b). The curved segments are arcs of circles of angle 35o and radii a = 12 cm and b = 6 cm. The straight segments are along radii. Find the magnetic field B at point P ( the center of curvature), assuming a current of 1 A clockwise in the circuit. (Positive out of page and negative into page) ONLY THE CURVED SEGMENTS CONTRIBUtE