A closely wound, circular coil with radius 2.50 cmcm has 740 turns. Part A What must the current in the coil be if the magnetic
1 answer:
Answer:
The current in the coil is 4.086 A
Explanation:
Given;
radius of the circular coil, R = 2.5 cm = 0.025 m
number of turns of the circular coil, N = 740 turns
magnetic field at the center of the coil, B = 0.076 T
The magnetic field at the center of the coil is given by;
where;
μ₀ is permeability of free space = 4 x 10⁻⁷ m/A
I is the current in the coil
R is radius of the coil
N is the number of turns of the coil
The current in the circular coil is given by
Therefore, the current in the coil is 4.086 A
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Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 × m
diameter d = 0.074 cm = 74 × m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W
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: