# Video: Physics Past Exam • 2017/2018 • Pack 1 • Question 16

Physics Past Exam • 2017/2018 • Pack 1 • Question 16

03:25

### Video Transcript

In the figure shown, a long, straight wire carrying a current 𝐼 one is placed tangentially to a circular ring of radius 𝑟 carrying a current 𝐼 two in the direction shown. A neutral point is formed at the center of the ring. Which of the following gives the ratio 𝐼 one to 𝐼 two and the direction of current 𝐼 one? a) 𝜋 upward, b) 𝜋 downward, c) one over 𝜋 upward, d) one over 𝜋 downward.

Looking at the figure, we can start off by calculating the direction of the magnetic field formed by current 𝐼 two as it moves around this circular ring. Using our right-hand rule, we know that 𝐼 two will form a magnetic field — we can call it 𝐵 two — that points into the page at the point 𝑟 at the center of the circle.

The problem statement tells us that a neutral point is formed at this center. That means the net magnetic field at that location is zero. So current 𝐼 one, which we’re told it runs through the long straight wire tangent to the circle, must create a magnetic field — we’ll call it 𝐵 one — that points out of the page at that location. At point 𝑟, 𝐵 one and 𝐵 two must be equal in magnitude and opposite in direction in order for a neutral point to form.

Based on this direction of 𝐵 one, once again using our right-hand rule, we see that 𝐼 one, the current in the vertical wire, must be moving upward. It’s current moving in that direction that will create a magnetic field at point 𝑟, pointing out of the page.

Knowing that 𝐼 one points upward, we can eliminate answer choices b and d from contention. They both involve 𝐼 one pointing downward. Knowing that at point 𝑟, 𝐵 one and 𝐵 two have equal magnitudes, we next wanna work back and solve for the ratio of the currents that generate those magnetic fields 𝐼 one to 𝐼 two.

If we look up the mathematical relationship for the magnetic field formed at the center of a current-carrying ring, we see that it’s equal to 𝜇 nought, the permeability of free space, times the current 𝐼 moving through the ring all divided by two times the ring’s radius. Applied to our scenario, we could say that the magnitude of 𝐵 two, the field formed at the center of the ring, is equal to 𝜇 nought times 𝐼 two over two times lowercase 𝑟.

If we next look up the relation solving for the magnetic field produced by a straight line of current-carrying wire, that’s equal to 𝜇 nought multiplied by the current in the wire all divided by two 𝜋 times the distance from the wire that the field is being calculated. We could express this in our terms as the magnitude of 𝐵 one is equal to 𝜇 nought times 𝐼 one all divided by two 𝜋 times 𝑟.

We’ve seen that because point 𝑟 is a neutral point, that means that the magnitude of 𝐵 two is equal to the magnitude of 𝐵 one. That being the case, we can write that 𝜇 nought 𝐼 two over two 𝑟 is equal to 𝜇 nought 𝐼 one over two 𝜋𝑟. We see that the factors of 𝜇 nought cancel out as to the factors of one over two 𝑟 in each side of this equation. 𝐼 two then is equal to 𝐼 one over 𝜋. And when we solve for the ratio 𝐼 one over 𝐼 two, we find that it is equal to 𝜋.

This tells us what our answer is. We want to choose choice a. The ratio 𝐼 one to 𝐼 two is equal to 𝜋 and the current 𝐼 one moves upward.

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