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Refer to the figure. A bar (L = 40 cm) moves on two frictionless rails in a regi

ID: 1521127 • Letter: R

Question

Refer to the figure. A bar (L = 40 cm) moves on two frictionless rails in a region where the magnetic field is uniform (B = 0.50directed into the paper. If v = 50 cm's and R= 80 m 2, what is the magnetic force on the moving bar? A straight section of a wire of length L = 80 cm lying along the x axis within a uniform magnetic field, B = 0.4 T perpendicular to the wire carries a current 1 = 5 A Find the magnitude of the magnetic force acting on the section of the wire. 9- A rectangular coil of dimensions a 4 cm and b - 6 cm carries current I = 5 A. A magnetic field B - 0.2 T is applied parallel to the plane of the coil. Find the magnitude of the magnetic dipole moment and the magnitude of the torque acting on the coil. 10-Two long parallel wires separated by 4.0 cm carry a current I] - 5 A and I2 = 8 A in opposite directions. What is the magnitude and direction of the magnetic field at point K at a distance d = 2 cm above the upper wire. (Magnetic field at a distance r outside of a long straight wire is given

Explanation / Answer

7) magnetic force   = (0.50 * 0.40 * 0.50/0.08) * 0.40 * 0.50

                              = 0.25 N

8)   magnetic force = I * L * B

                             = 0.8 * 0.4 * 5 = 1.6 N

9) manetic dipole moment = 5 * 0.04 * 0.06 = 0.012 A-m2

     torque = 0.2 * 0.012   = 2.4 * 10-3 N.m

10) a) magnetic field =   2 * 10-7 * 5/0.02   -   2 * 10-7 * 8/0.06

                               =   2.33 * 10-5 T                  ( Direction = out of plane of paper)

     b)   Direction = out of plane of paper

11) magnitude of magnetic field = sqrt(32 + 42) * 10-6

                                                                        = 5 * 10-6 T

12)   magnitude of magnetic field inside solenoid = 4 * 3.14 * 10-7 * 8000 * 0.07

                                                                            = 7.033 * 10-4 T

13) induced current = 20 * 50 * 10-4 * 4 /(2 * 0.40)

                               = 0.5 A

14)    current induced = 2 * 10-7 * 1.5 * ln(6/5) * 100/(2 * 10-3)

                                  = 2.734 * 10-3 A