Question

Set up for problem is given in the picture. a-c are good. Please help with part d (not in picture)

N=1500 7. [***] A square coil is at the center of an ideal solenoid as shown. The coil has 6 turns with sides 0.050 m long. The solenoid is 0.300 m long, has 1500 turns of wire and a radius of 0.15 m. a. A current of 13.0 A flows in a counter-clockwise direction around the solenoid. Calculate the magnetic field in the solenoid and indicate its direction on the diagram. b. The current in the solenoid increases to 22.0 A in a time of 0.025 s. If the square coil has a total resistance of 10.0 2, determine the induced current in the square coil and indicate its direction on the diagram. (Hint find the new B-field) c. Consider a single wire loop going around the outside of the solenoid with a radius of 0.25 m (centered on the solenoid's axis). Given the same change in current from part b, find the magnitude of the induced emf in the loop. (Assume I outside the solenoid is zero)

d) Next, the current in the solenoid is varied in time as the function, ?(?) = ?0e−t/τ, where ?0 = 13.0 A
and τ = 5.24 μs. Determine the induced current in the coil at t = 4.00 μs. *Answer is 10.9 A, must show all work

Answer 1

Page No. → solution of a → we are given L(E)=le e-te - I Lo = 13,0 A h oc=5.24 lls. to 4.00 US • we want to find I2 a We also have al=6 = Number of furns en square core N = 1500 3 = Number of luons in solenoid I lg = 0.0som = Length of side in square coil l = 0.300m = length of solenoid 8 = 0.15 m = Radius of solenoid A Tir2 = 0.07065m² = Area of solenoid A2 = ls² = 0,0025 m² 2 Area Area of of square coil Ho - 4 10-7- Permeability of space Let's find in south - - 14/5.24) I (4) - 13 x l 6.06 A = 1 - | Let's find Bid magnetic field in solenoid tai Ballon I, = 4 TT *10*7 x N: 26.06_ U M DAN = 4T X.10+* * 1500 x 6.06 0.3 B. - 38.07x10-T → that n is the number of turns Please note per length

Page No. Date Now let's find o fluse linkage with solenoid : 0,=N, B, A, cosa I where, = 1500 x 38,07x 10-3_ W, = Number of teenus dan setened x 0.07065 x cas O Bi - Magnetic field in solenoid A = Area of solenoid 6. = 4.03447 Wbm? Lo=Angle between B, and A, Y | = 0 (°: parallell Qi=4.03 wb m² Magnetic field Bi is almost similar to the magnetic field B2 for square coll enside the solenoid Therefore, Bi=B, Magultic field of solenoiel B = 35,07x10-3 T Magnetic field of quare coil B2 = 36.07 x1013 T → Let's fing 02, fluse linkage with square coil 1 42 = A, B, 4, Case = 6 x 38.07x1013x 0,0025 12 = 0.57105 x 10-3 wb m² . I Let's find mutual enduefance M between solenoid and square core • M= N2 02 - 6x0.57105 *10-3 I 6.06 M = 0.5654 *103 H

Page No. Date Inally i t let's find the value of L2 I2 = Ņe di M I2 = 1500 x 4.03 10.7 MA 0.5654 X10-3 - - I2 = 10,7uA