Question

SECTION: 1 (8am) 2H08mJu cular flat coil A of radius 1 cm lies inside the center of a long solenoid bout an axis that is along its own diameter. That axis also lies perpendicular to at coil A of radius 1 cm lies inside the center of a long solenoid B with a axis turn density of 10 turns/cm. Coil A is free to rotate a the solenoidal axis and passes through the center of the solenoid. A current of 5 A is made to to vacuum permeability is o 4x 107 T m/A. flow through both A and B. The a) In the cut-away diagram provided, assign the solenoid current directions such that its magnetic field points left to right, and the flat coil current is counter-clockwise. At locations 1 thru 4, specify the direction of the force at each location. Include the magnetic moment vector and torque vector on the diagram. (5 points) Solenoid axis b) Calculate the magnitude of the magnetic moment of coil A. (5 points) c) What is the maximum torque that A could experience about its axis (5 points) What angle would A's magnetic moment vector make with the solenoidal axis such that the torque were only half that of part c)? (5 points) d) e) As the loop begins to rotate (from rest) in response to the torque, a new current component will be induced in the loop by the external field. Will the induced current in the loop be in the same direction as the original current, or in the opposite direction? State your answer, and explain in words how to use principles from the course to support that direction for induced current. Try to y-step in your explanation. Hint: What direction does the magnetic moment point? In which direction will the loop rotate? (5 points)

Answer 1

(a) The force acting the coil is given as, Here, , is the current in the coil The magnetic moment is in the direction of the normal to the coil and it is given as, m NI,A Here, N is the number of turns in the coil and A is the area of the coil. The torque acting on the coil is given as The following figure shows the direction of solenoid current I,. From right hand rule, force at point 1 is out of the page and it is into the page at point 3. At point 2 and 4, the direction of current is antiparallel and parallel to the field. Therefore, force at point 2 and 4 is zero The direction of magnetic moment (m) is out of the page and the torque (r) acting on the coil is in the upward direction y 2 T Solenoid axis 3 7 m B 4 /1X XXXXXXX Coil axis L

(b) The magnitude of the magnetic moment of coil A can be calculated as follows: т 3D N,A NI, пr; Here, is the radius of the coil 2 Plug in the values in the above equation and calculated the magnitude of the magnetic moment of coil A m(50)(5 A)7(0.01 m) - 2 -0.0785 A m (c) The maximum torque can be calculated by using the following relation: = mB sin e Here, B is the magnetic field due to the solenoid The magnetic field inside the solenoid is given as, Here, n is the number if turns per unit length and t is the vacuum permeability

For maximum torque, 0 90°. Therefore, the maximum torque experienced by the coil is, T m(Hnl^)sin90° - "nuти — (0.0785 A m2)(4x10-7 T-m/A 10 turns/cm 100 cm 1 m |(5 A) =4.9348x10 N-m 4.93x10 N-m (d) The angle at which the torque is half of its initial value can be calculated as follows: r'= mB sin т(и,п1,)sin ф 2 =sin 2тu,nl, Plug in the values in the above equation and calculate the angle 4.9348 x10 N m sin 100 cm (5 A) 2(0.0785 A m2(4xx10-7 T-m/A) 10 turns/cm 1 m 47T = 30°