Question

12999 3. What is the magnetic field (magnitude and direction) at each of the positions a-c in the figure to the side? What force will each wire experience (per unit length)? 10A 120 cm 4. Consider a current-carrying wire that has been twisted to form a half circle of radius r. Use the Biot-Savart Law to set up an integral for the magnetic field at the center of the circle. As usual, do not solve the integral, but clearly set it up. 1.0 cm Integration 5. What is the line integral of the magnetic field between points i and f in the figure to the side? (Determine a numerical value.) Could you use this path with Ampere's Law to determine an equation for the magnetic field generated by a current-carrying wire? Why or why not?

Answer 1

Solution (3):

Magnetic field at point at point (a):

Magnetic field at point a, due to the current flowing right dirrection is given by

$ B1.di = uolenc - 2hr B = Hol > Bi = 0 2^ - Bi = 2^T *.. r = 2.0cm = 2x 10°m I= 10.04 .B 4TX 10-7 x 10 | 20 x2x 10-2 = B = 1.0 x 10-16

Similarly the magnetic field at point a, due to the current flowing left dirrection is given by

| + B.lt = Polene →2rr B. = lol > B = of * B. (-4) r = 6.0cm =6×10-Pm I = 10.04 了 B = -1 40 x 10-7 x 10. 20 x6x10-22 B = -0.33 x 10-16

So, net magnetic field at point a, due to the current carrying wires is given by

(1)

B = B1 + B2 Ba = (1.0 – 0.330 x 1024 B = 0.66 x 10-4 =

Magnetic field at point at point (b):

Magnetic field at point b, due to the current flowing right dirrection is given by

* > B3 = of $ 33.d1 = polenc = 2^r B3 = Hol -20 = B = 60 (-) F=2.0cm = 2x 10- Pm I = 10.0.4 *.. B 4TX 10 x 10 2T x2x 10-2 B = -1.0 x 10-0

Similarly the magnetic field at point b, due to the current flowing left dirrection is given by

$ Ba.di = uolenc → 2r B4 = lol > BI = 0 = B = (一) r = 2.0cm = 2x 10-Pm I = 10.04 4 x 10-7x10. 20x2x 10-2 B4 = -1.0 x 10-0

So, net magnetic field at point b, due to the current carrying wires is given by

(2)

Bo = B: +34 Bo = (-1.0, 1.0) x 10-40- B = -2.0 x 10-40 By = 2.0 x 10-40-0)

Magnetic field at point at point (c):

Magnetic field at point c, due to the current flowing right dirrection is given by

+ Badt = polance → 2r B = lol fff.. > Bs = of → s = {(-4) r = 6.0crm = 6x 10-2m I=10.0.4 a 4T x 10-7x10。 20 x6x10-22 B = -0.33 x 10-16

Similarly the magnetic field at point c, due to the current flowing left dirrection is given by

$ B.di = polenc 2r Be = HoI fff.. > B = Wo/ B = 0! | 2TT r = 2.0cm = 2x 10-2mm I = 10.0.4 R 4TX 10-7 x 10 | x2x10-29 = B = 1.0 x 10-16

So, net magnetic field at point c, due to the current carrying wires is given by

(3)

B = B5 + B6 → B = (-0.33 + 1.0) x 10-40- Be=0.66 x 10-40

Now, calculation force per unit length on each wire:

Since current in each wire is same and in opposite direction. So, the force per unit length is attractive in nature and equal in magnitude.

Magnetic field on one wire due to the other is given by

$ B.di = Molenc 2 rb = pol BHOZ D 2 = 4 x 10-2m r = 4.0cm I = 10.04 B-47 10-7 x 10 B= 27 4 x 10-2 =B=0.5 x 10-4

and magnetic force per unit length on one wire due to the other is given by

F f= idl →f= I Вsing 1:0 = 90° f=IB

Putting the values of I and B, we get

(4)

f = 10 x 0.5 x 10-4 =f=5.0 10-4n