Question

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

a) What is the magnitude of the magnetic field at the center of the coil?
b) What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 9.50 cm from its center?

Answer 1

Part A is B= 5.65×10-3

Answer 2

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

`dB = (\mu_0)/(4\pi) \frac{id\vec{l} * \hat{r}}{r^2}`

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.

`dB = (\mu_0)/(4\pi) \frac{id\vec{l}}{r^2}`

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:

`B = \int (\mu_0)/(4\pi) (ids)/(R^2)`

Taking out constants from the integral:

`B =(\mu_0 i)/(4\pi R^2) \int ds`

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

`B =(\mu_0 i)/(4\pi R^2) \int\limits^(2\pi R)_0 \, ds`

Evaluate:

`B =(\mu_0 i)/(4\pi R^2) (2\pi R- 0) = (\mu_0 i)/(2R)`

Plugging in our givens to solve for the magnetic field strength of one loop:

`B = ((4\pi *10^(-7)) (0.460))/(2(0.0215)) = 1.3443 \mu T`

Multiply by the number of loops to find the total magnetic field:

`B_T = N B = 0.00631 = \boxed{6.318 mT}`

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:

`dB = (\mu_0)/(4\pi) \frac{id\vec{l} * \hat{r}}{r^2}`

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.

`dB = (\mu_0)/(4\pi) \frac{id\vec{l} sin\theta}{r^2}`

Using the diagram, if 'z' is the point's height from the center:

`r = √(z^2 + R^2 )\\\\sin\phi = (R)/(√(z^2 + R^2))`

Substituting this into our expression:

`dB = (\mu_0)/(4\pi) \frac{id\vec{l}}{(√(z^2 + R^2))^2} }((R)/(√(z^2 + R^2)))\\\\dB = (\mu_0)/(4\pi) \frac{iRd\vec{l}}{(z^2 + R^2)^(3)/(2)} }`

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:

`B = (\mu_0 iR)/(4\pi (z^2 + R^2)^(3)/(2))} \int\limits^(2\pi R)_0, ds`

Evaluate:

`B = (\mu_0 iR)/(4\pi (z^2 + R^2)^(3)/(2))} (2\pi R)\\\\B = (\mu_0 iR^2)/(2 (z^2 + R^2)^(3)/(2))}`

Multiplying by the number of loops:

`B_T= (\mu_0 N iR^2)/(2 (z^2 + R^2)^(3)/(2))}`

Plug in the given values:

`B_T= ((4\pi *10^(-7)) (470) (0.460)(0.0215)^2)/(2 ((0.095)^2 + (0.0215)^2)^(3)/(2))} \\\\ = 0.00006795 = \boxed{67.952 \mu T}`