When you are driving down the road you notice a large wire running parallel to t

Question

When you are driving down the road you notice a large wire running parallel to the surface on the right hand side of the road. The wire is 7.50 m above ground. If the wire was carrying a DC current (which it is not) in the opposite direction you are traveling, how much current would be required to produce a magnetic field at the ground with a magnitude equal to that of the Earth's magnitude field at that point which is known to be 0.625 gauss = 6.25 times 10^3 T? What direction this filed be pointed at a location directly under the wire? Three wires are arranged in a line as shown below. All wires carry current perpendicular to the plane of the page. The wires located at x = 0.300 m and +0.700 m both have a current of 3.00 A going into the page, the wire at x = 0.0 m carries a current of 1.75 A out of the page. What is the magnetic field (magnitude and direction) at position P at x = +0.200 m?

Explanation / Answer

P1)

Magnetic field due to the current (I) on the surface ,

B = u*I/(2*pi*d)

where d = 7.5 m

u = permeability of free space = 1.26*10^-6

So, B = 1.26*10^-6*(I)/(2*pi*7.5) = 6.25*10^-5

So, I = 2337.5 A <---------answer

P2)

Magnetic field due to 3A current(on the far left) at P:

B1 = u*I/(2*pi*d)

here I = 3A,

u = permeability of free space = 1.26*10^-6

d = distance between 3A and pont P = 0.3 + 0.2 = 0.5 m

So, B1 = 1.26*10^-6*(3)/(2*pi*0.5)

or , B1 = 1.2*10^-6 T (directed down)

Similalrly magnetic field due to 1.75 A current,

B2 = 1.26*10^-6*(1.75)/(2*pi*0.2)

= 1.75*10^-6 T (directed up)

and ,

magnetic field due to 3A on the right:

B3 = 1.26*10^-6*(3)/(2*pi*0.5)

= 1.2*10^-6 T (directed up)

So, net magnetic field due to all the currents:

Bnet = -B1 + B2 + B3 <--------- negative sign for B1 is due to direction towards down

So, Bnet = -1.2*10^-6 + 1.75*10^-6 + 1.2*10^-6

So, Bnet = 1.75*10^-6 T (directed up)<----------answer

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.