A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she
1 answer:
You might be interested in
Hi there!
a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.
dB = Differential Magnetic field element
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
R = radius of loop (2.15 cm = 0.0215 m)
i = Current in loop (0.460 A)
For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.
Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
Taking out constants from the integral:
Since we are integrating around an entire circle, we are integrating from 0 to 2π.
Evaluate:
Plugging in our givens to solve for the magnetic field strength of one loop:
Multiply by the number of loops to find the total magnetic field:
b)
Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.
Using the diagram, if 'z' is the point's height from the center:
Substituting this into our expression:
Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
Evaluate:
Multiplying by the number of loops:
Plug in the given values:
Answer:
18 Ω
Explanation:
As K and F are at the same voltage, we can redraw the diagram as in figure 2
Series resistances add directly, so we get figure 3
Adding parallel resistances gets us to figure 4
Now we can move two 6Ω resistances for clarification in figure 5
As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6
Add series resistances to get to figure 7
Add parallel resistances to get to figure 8
Add series resistances to get to figure 9
