Point between two wires at which the flux density is zero

AI Thread Summary
The discussion focuses on calculating the magnetic flux density between two parallel wires carrying different currents. When both currents are in the same direction, the flux density midway is 3.2 * 10^-4 T, while it is 4.8 * 10^-4 T when the currents are in opposite directions. The point where the flux density is zero is located 2.5 cm from the wire carrying 30 A. The calculations involve using the formula for magnetic flux density and considering the distances from each wire. Accurate unit conversion is emphasized to avoid confusion in the results.
moenste
Messages
711
Reaction score
12

Homework Statement


A long wire (X) carrying a current of 30 A is placed parallel to, and 3.0 cm away from, a similar wire (Y) carrying a current of 6.0 A. What is the flux density midway between the wires: (a) when the currents are in the same direction, (b) when they are in opposite directions? (c) When the currents are in the same direction there is a point somewhere between X and Y at which the flux density is zero. How far from X is this point? (μ0 = 4 π * 10 -7 H m-1.)

Answers: (a) 3.2 * 10-4 T, (b) 4.8 * 10-4 T, (c) 2.5 cm

2. The attempt at a solution
(a) B = μ0 I / 2 π a = [4 π * 10 -7 * (30 - 6)] / 2 π * (0.03 / 2) = 3.2 * 10-4 T. I is (30 - 6) since the currents are in the same direction and 0.03 is divided by two because we need the flux density midway between the wires.

(b) Same formula but 30 + 6, since the currents are in opposite directions = 4.8 * 10-4 T.

(c) Regarding this part I don't know. I used the abovementioned formula and got: [4 π * 10 -7 * (30 - 6)] / 2 π * a = 0 → 7.2 * 10-6 a-1 = 0. And that's as far as I got.
 
Physics news on Phys.org
The way you have it written it looks like you expect the point to be equidistant from both wires. Does that make sense to you, knowing the wires have different currents?
 
  • Like
Likes moenste
The Buttered Cat said:
The way you have it written it looks like you expect the point to be equidistant from both wires. Does that make sense to you, knowing the wires have different currents?
Hm, I think I got it:
μ0 IX / 2 π aX = μ0 IY / 2 π aY
μ0 IX 2 π aY = μ0 IY 2 π aX
30 aY = 6 aX
5 aY = aX

We know that aX + aY = 0.03 m

Since we need aX: aY = 0.2 aX → aX + 0.2 aX = 0.03 → 1.2 aX = 0.03 → aX = 0.025 cm or 2.5 m.
 
moenste said:
aX = 0.025 cm or 2.5 m.

Be careful with your units there, but otherwise you are exactly right.
 
  • Like
Likes moenste
The Buttered Cat said:
Be careful with your units there, but otherwise you are exactly right.
Oh yes, it's the other way around :). 0.025 m and 2.5 cm.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top