A 5.45-g combustible sample is burned in a calorimeter. the heat generated changes the temperature of 555 g of water from 20.5°c
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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: