A very large loop of metal wire with radius 1 meter is driven with a linearly increasing current at a rate of 200 amps/second. A very small metal wire loop with radius 5 centimeter is positioned a small distance away with its center on the same axis (the loops are coaxial). The small loop experiences an induced emf of 983 nano-volts. What is the separation of the loops in meters? Note that a subtraction step in the solution makes it sensitive to significant figures. Keep at least four figures in your calculation
To find the separation of the loops, we can use Faraday's Law of electromagnetic induction, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop.
The magnetic flux through a loop with radius r due to a current I is given by:
Φ = μ₀ * I * π * r²
where μ₀ is the permeability of free space (4π × 10^-7 T·m/A).
In this case, the large loop is driven with a linearly increasing current, meaning the current changes with time. We need to calculate the rate of change of magnetic flux to find the separation between the loops.
The rate of change of magnetic flux (dΦ/dt) is given by:
dΦ/dt = μ₀ * (dI/dt) * π * r²
Given that the current is increasing at a rate of 200 A/s, we have:
dI/dt = 200 A/s
The radius of the large loop is 1 meter, so its magnetic flux is:
Φ = μ₀ * I * π * (1 m)²
= μ₀ * I * π
Now, let's calculate the rate of change of magnetic flux:
dΦ/dt = μ₀ * (dI/dt) * π * (1 m)²
= μ₀ * (200 A/s) * π
Finally, we need to calculate the separation between the loops. The induced emf in the small loop is given by:
emf = - (dΦ/dt) * N
where N is the number of turns in the small loop. Since we have only one small loop, N = 1.
Plugging in the values:
983 nV = - (μ₀ * (200 A/s) * π) * 1
Let's solve for μ₀:
983 nV = - (200 A/s) * (4π × 10^-7 T·m/A) * π
μ₀ = - (983 nV) / [(200 A/s) * (4π × 10^-7 T·m/A) * π]
Now, we can substitute the value of μ₀ into Faraday's Law to find the separation between the loops:
μ₀ = 4π × 10^-7 T·m/A
Substituting and simplifying:
4π × 10^-7 T·m/A = - (983 nV) / [(200 A/s) * (4π × 10^-7 T·m/A) * π]
Solving for the separation of the loops:
Separation = - (983 nV) / [(200 A/s) * (4π × 10^-7 T·m/A) * π]
≈ -0.02027575911 meters
Keeping four significant figures, the separation between the loops is approximately -0.02028 meters. Note that the negative sign indicates that the small loop is positioned on the opposite side of the large loop.