The two wires shown in the figure below carry currents of 5.00 A in opposite dir

Question

The two wires shown in the figure below carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field at the following locations: at a point midway between the wires 4e-5 T into the page at point P1, 10.0 cm to the right of the wire on the right 2.666e-5 Your answer differs from the correct answer by 10% to 100%. T to the right at point p2, 20.0 cm to the left of the wire on the left 2.5e-5 Your answer differs from the correct answer by orders of magnitude. T into the page Need Help? Road It Chat About It Submit New Answers To Question 5 Save Work

Explanation / Answer

Current in the wires is I = 5 A Now magnetic field due to left wire at  midwaybetween the wires is BL =0I/2r 0 = 4*10-7H/m r = half of distance betwen the wires = 10 cm /2 = 5 cm = 0.05 m Now according to amperes right hand thumb rule magnetic fielddue to left wire is into the page. Then BL = 0I/2r =(4*10-7H/m )*5A/2*0.05 m =2*10-5T Now magnetic field due to right wire at  midwaybetween the wires is BR =0I/2r 0 = 4*10-7H/m r = half of distance betwen the wires = 10 cm /2 = 5 cm = 0.05 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into the page. Then BR = 0I/2r =(4*10-7H/m )*5A/2*0.05 m =2*10-5T Then net field due to wires  at a point midwaybetween the wires is B =  BL+  BR                                                                                                         = 4*10-5T b) Now magnetic field due to left wire at  point P1 is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P1 is BR =0I/2r2 0 = 4*10-7H/m r2 == 10 cm = 0.1 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is OUT OF the page Then BR=0I/2r1 =(4*10-7H/m )*5A/2*0.1 m =10*10-6T = 1*10-5T Then net field due to wires at point P1 is B =BR -BL = 1*10-5T-  0.5*10-5T                                                                                        =  0.5*10-5T (out of page) c ) Now magnetic field due to left wire at  pointP2   is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is out of the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) 0 = 4*10-7H/m r = half of distance betwen the wires = 10 cm /2 = 5 cm = 0.05 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into the page. Then BR = 0I/2r =(4*10-7H/m )*5A/2*0.05 m =2*10-5T Then net field due to wires  at a point midwaybetween the wires is B =  BL+  BR                                                                                                         = 4*10-5T b) Now magnetic field due to left wire at  point P1 is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into the page Now magnetic field due to left wire at  point P1 is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P1 is BR =0I/2r2 0 = 4*10-7H/m r2 == 10 cm = 0.1 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is OUT OF the page Then BR=0I/2r1 =(4*10-7H/m )*5A/2*0.1 m =10*10-6T = 1*10-5T Then net field due to wires at point P1 is B =BR -BL = 1*10-5T-  0.5*10-5T                                                                                        =  0.5*10-5T (out of page) c ) Now magnetic field due to left wire at  pointP2   is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is out of the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P1 is BR =0I/2r2 0 = 4*10-7H/m r2 == 10 cm = 0.1 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is OUT OF the page Then BR=0I/2r1 =(4*10-7H/m )*5A/2*0.1 m =10*10-6T = 1*10-5T Then net field due to wires at point P1 is B =BR -BL = 1*10-5T-  0.5*10-5T                                                                                        =  0.5*10-5T (out of page) c ) Now magnetic field due to left wire at  pointP2   is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is out of the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Now magnetic field due to right wireat  point P1 is BR =0I/2r2 0 = 4*10-7H/m r2 == 10 cm = 0.1 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is OUT OF the page Then BR=0I/2r1 =(4*10-7H/m )*5A/2*0.1 m =10*10-6T = 1*10-5T Then net field due to wires at point P1 is B =BR -BL = 1*10-5T-  0.5*10-5T                                                                                        =  0.5*10-5T (out of page) c ) Now magnetic field due to left wire at  pointP2   is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is out of the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Then BR=0I/2r1 =(4*10-7H/m )*5A/2*0.1 m =10*10-6T = 1*10-5T Then net field due to wires at point P1 is B =BR -BL = 1*10-5T-  0.5*10-5T                                                                                        =  0.5*10-5T (out of page) c ) Now magnetic field due to left wire at  pointP2   is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is out of the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Now magnetic field due to left wire at  pointP2   is BL =0I/2r1 0 = 4*10-7H/m r1 == 20 cm = 0.2 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is out of the page Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Then BL =0I/2r1 =(4*10-7H/m )*5A/2*0.2 m =5*10-6T = 0.5*10-5T Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Now magnetic field due to right wireat  point P2   is BR =0I/2r2 0 = 4*10-7H/m r2 == 30 cm = 0.3 m Now according to amperes right hand thumb rule magnetic fielddue to right wire is into  the page Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Then BR=0I/2r2 =(4*10-7H/m )*5A/2*0.3 m =3.333*10-6T = 0.333*10-5T Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page) Then net field due to wires at point P2 is B =BL -BR = 0.5*10-5T-  0.333*10-5T                                                                                        =  0.167*10-5T (out of page)

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