Please show the detailed solution for part 15,16,17 and 18. Please give detailed answer, Full Question is given in the images attached, Thanks! Ampere's Law was used to show that the magnitude of the magnetic field a distance R from a long, thin wirecarrying current I is given by:В27RILeftIrightA right hand rule can be used to determine the direction of thismagnetic field.dACTwo parallel wires, separated by a distance of 20 cm, are carrying currents. Thelefthand current, Iet, Wire carries a current ILet = 15 A in the direction shown andthe other carries a current Iright = 25A current as shown. Point C is a distance of10 cm to the right of the Iright current and Point A is halfway between the twowires. (Show your work below and on the following page. Be sure to label towhich section the work belongs).Color the left hand wire red and color the right hand wire green. Then, on the diagram, usingcorrespondingly colored ® and • to represent “into" and “out of" the page, respectively, show themagnetic field directions for each wire in all the regions left, right, and in-between the wires.10. What is the magnitude of the B-field at Point C due to Iright? Ans: CА. 1.66х10-5TD. 4.00x10$TB. 2.00x10ST| C. 5.00x10-$T11. What is the magnitude of the B-Field at Point C due to ILeh?B. 2.00x10³TAns: A|A. 1.00x10-$TC. 3.00x10-³TD. 4.00x10-°T12. What is the total magnetic field at Point C (both magnitude and direction must be correct for credit)?Ans: Ca. 1.34x10$ Tb. 7.00x10$ Tc. 6.00x105 Td. 0 T13. What is the DIRECTION of the total magnetic field at Point C? Ans: Aa. Into pageb. Leftc. Rightd. Out of page 15. What is the direction of that force on the proton? Ans: Bc. Intob. Downa. Upd. Left16. If the currents remain the same as found in above, what will be the magnitude of the magnetic fieldfor Point A (remember, this will be the result of both wires (don't forget to take account thedirections of each B-field from each wire)? Ans: BА. 1.66х105TB. 2.00x10-°T|С. 3.00х105TD. 4.00x10-ST17. What will be the direction of the B-field at Point A? Ans: Ba. Into the pageb. Out of the page18. If a third current carrying wire is laid parallel to the original wires carrying a 5A current downwardand passing through Point A, what will be the force per unit length on that wire (remember, there is aformula for the force on a current-carrying wire immersed in a magnetic field)? Ans: AA. 1.00x10“NB. 2.00x10“N|С. 3.00х10-ND. 4.00x 10“N

You can find it by pointing your right thumb in the direction of the current in the wire and curling your fingers. The magnetic field around the wire will cause your fingers to curl in the same direction.15 Calculate the direction of force on the proton. F=qv→×B→ Here, velocity v→ is left and magnetic field B→ into the page. Thus, the direction of force on the proton is down. Hence, option B is the correct answer.16 As given, Current on left wire, Ileft=15 A Current on right wire, Iright=25 A Distance between two wires, d=10 cm=0.1 m Using right hand rule: Calculate magnetic field due to left wire. Bleft=μolleft2πd=4π×10-7×lleft2πd=2×10-7×150.1=3×10-5 T This magnetic field into the page. Determine magnetic field due to right wire. Bright=μolright2πd=4π×10-7×lright2πd=2×10-7×250.1=5×10-5 T This magnetic field out of the page. Calculate the net magnetic field: B=Bright-Bleft=5-3×10-5=2×10-5 Hence, the magnetic field at point A is 2×10-5. Thus option B is the correct answer.17 The direction of magnetic field at point A is out of the page. Hence, option B is the correct answer. 18 As given, Current, I=5 A Force per unit length due to left wire: F=ILBFL=IB=5×2×10-5=10×10-5 Nm=1×10-4 Nm Hence, the force per unit length is FL=1×10-4 Nm. Therefore, option A is the correct answer.