A piece of copper wire has a resistance per unit length of 6.45*10-3 /m. The wire is wound into a thin, flat coil of many turns that has a radius of 0.140 m. The ends of the wire are connected to a 12.0 V battery. Find the magnetic field strength at the center of the coil.
First calculate the resistance of the wire R, using the length L and the resistance per unit length.
You will have to leave L as an unknown. It will cancel out later.
Then calculate the current I using Ohm's law.
You will also need the number of turns of wire, N, from
N = L/(2 pi R)
For the B-field in the middle of the coil,
B = mu N I /(2 R)
To find the magnetic field strength at the center of the coil, we can use Ampere's Law. Ampere's Law states that the line integral of the magnetic field around a closed path is equal to the product of the permeability of free space (μ₀) and the current flowing through the enclosed area.
The formula for the magnetic field strength inside a coil is given by:
B = (μ₀ * n * I) / (2 * r)
where:
B = magnetic field strength,
μ₀ = permeability of free space (4π * 10^-7 T*m/A),
n = number of turns of the coil,
I = current flowing through the coil (in Amperes), and
r = radius of the coil.
First, let's find the current flowing through the coil. We can use Ohm's Law:
V = I * R
where:
V = voltage (12.0 V in this case),
I = current, and
R = resistance per unit length * length of the wire (assuming the wire length is equal to the coil circumference).
The length of the wire (l) is given by the circumference of the coil:
l = 2π * r
Substituting the values we have:
R = 6.45 * 10^-3 Ω/m * (2π * 0.140 m)
R ≈ 0.018Ω
So, using Ohm's Law:
12.0 V = I * 0.018Ω
I ≈ 12.0 V / 0.018Ω
I ≈ 666.7 A
Now, let's substitute the values into the magnetic field formula:
B = (4π * 10^-7 T*m/A) * n * 666.7 A / (2 * 0.140 m)
B ≈ (4π * 10^-7) * n * 666.7 / (2 * 0.140) T
B ≈ 9.55 * 10^-6 * n T
Therefore, the magnetic field strength at the center of the coil is approximately 9.55 * 10^-6 times the number of turns (n) in Tesla (T).
To find the magnetic field strength at the center of the coil, we can use Ampere's Law. Ampere's Law states that the integral of the magnetic field B dot ds along a closed loop is equal to μ₀ times the net enclosed current I.
In this case, the net enclosed current is the total current passing through the coil. Since the ends of the wire are connected to a 12.0 V battery, we can use Ohm's Law to find the current flowing through the wire.
Ohm's Law states that the current I is equal to the voltage V divided by the resistance R. The resistance of the wire can be found using the resistance per unit length and the length of the wire.
Now, let's calculate the total resistance of the wire. Since the wire is wound into a thin, flat coil, the total length of the wire can be found using the formula for the circumference of a circle.
Circumference = 2πr = 2π(0.140 m) = 0.88 m
Based on the given resistance per unit length of 6.45 x 10^-3 Ω/m, we can calculate the total resistance R.
R = resistance per unit length x length of wire = (6.45 x 10^-3 Ω/m) x (0.88 m) = 5.68 x 10^-3 Ω
Now, let's calculate the current flowing through the wire.
I = V / R = 12.0 V / 5.68 x 10^-3 Ω = 2.11 A
We now have the value of the current I. The next step is to calculate the magnetic field at the center of the coil using Ampere's Law.
According to Ampere's Law, the line integral of the magnetic field B dot ds along a closed loop is equal to μ₀ times the net enclosed current I. For a circular loop of radius R centered at the origin, the magnetic field B at the center of the loop is given by:
B = (μ₀ x I) / (2πR)
In this case, the radius of the coil is 0.140 m and the current I is 2.11 A.
Let's calculate the magnetic field B at the center of the coil.
B = (μ₀ x I) / (2πR) = (4π x 10^-7 T·m/A x 2.11 A) / (2π x 0.140 m)
= (4π x 10^-7 T·m/A x 2.11 A) / (0.280 m)
= 3.00 x 10^-6 T
Therefore, the magnetic field strength at the center of the coil is 3.00 x 10^-6 T.