Question

I only need help with questions 4,5,6,7, and 8. thanks

Magnetic field and Magnetic Force due to a (long, straight) current PHYS 181 - in class problem set An infinitely long conductor carrying current is bent at a right angle as shown in Figure 1. Point P is located a distance b from the corner of the wire. Only one section of this current contributes to the magnetic field at pt. P. Why? The general formula (derived from the Biot-Savart Law) for the magnitude of the magnetic field a distance a away from a thin, straight conductor is: B ! (sin e, - sin 0,) For this problem, what are the values for the following variables: a = b 0,0 0 ,_ 90 is alwe should be negatie...why?) (1.) 0.75 point) ir b = 1.00 cm and I = 6.00 A, what is the magnitude of the magnetic field at point P, in micro-Tesla)? R= м. Т . IT ҳ10 е UT .ol -4 TT 10 _-. OO 4 TT (2.) [0.25 point] What is the direction of the magnetic field at point P? (ind direction sing of the band ruley into the page y, very long conductor is located a distance 2b to the right of shown in Figure 2). and carries a current of 31 toward the top of as shown (3.) 0.5 point] What is the direction of the magnetic field due to just the second straight current at point P? out of the page 9 11 point] What is the net magnitude of the magnetic field at point P due to both current-carrying conductors, in je (5.) 10.5 point] What is the direction in What is the direction of the net magnetic field (due to both current-carrying conductors) at point P?

Answer 1

1)

Magnetic field at point P due to first wire is

μο11 μο, Β1 = . (sin 0° – sin 90°) = 4πι 4π * 10-7 και 6.00 = 6 και 10-5T 4π(0.01) 4πd

2)

Direction of magnetic field is into the page.

3)

Direction of magnetic field due to second wire is   out of the page.

4)

Magnetic field at P due to second long wire is  

Bz = /2 (sin Oº – sin 180°) = 4:03. (2) = 45,* 47 * 10-7* 18.0 27(2*0.01) = 18 * 10-51 4r

( out of page is taken positive, into the page is taken negative)

net magnetic field at point P due to two wires is

B = -B1 + B2 = -6 * 10-5 + 18 * 10-5 = 12 * 10-5T

5)

Net magnetic field is positive , so it is out of the page.

Direction of net magnetic field is out of the page.

6)

11=1 = 6.00 A, 12 = 91 = 54.0 A

Now the two wires are long, and are separated by a distance

T = 3b = 0,03 m

Force per unit length between two wires is

μο1112 2πι L

Magnitude of magnetic force per unit length on conductor 1 is

Fi Holl247 * 10-7* 6.00 * 54.0 L2r 27(0.03) = 2160 * 10-6 N/m = 2160 u N/m

7)

Magnitude of magnetic force per unit length on conductor 2 is

F2 MolI2 41 * 10-'* 6.00 * 54.0 L2=1 27(0.03) = 2160 * 10-6 N/m = 2160 u N/m

8)

Direction of magnetic force on conductor 2 is towards left.

Direction of magnetic force on conductor 1 is towards right.​​​​​​​