Question

,a circular loop carrying a current of 40A produce a field of 3*10^8 T at the center what should be the current in the straight conductor so that it produce the same field at a distance equal to the radius if the loop ?​

Answer 1

Hi there!

We can begin by deriving the magnetic field strength for a circular loop at its center.

Using Biot-Savart's Law:

`dB = (\mu_0)/(4\pi )(idl * \^r)/(r^2)`

μ₀ = Permeability of free space (Tm/A)

dl = Differential length element

r = radius (m)

We can rather use 'ds' to represent a differential arc length since we are finding the magnetic field for a circular loop.

We must begin by dealing with the cross-product. Luckily, this derivation is simple since the radius vector is ALWAYS perpendicular to the path of integration along the loop's circumference. (ONLY if we are finding the field for the direct center.) Thus, since sin(90) = 1, we can get rid of the cross-product.

Therefore:

`dB = (\mu_0)/(4\pi )(ids)/(r^2)`

Now, we can integrate with respect to ds.

`= (\mu_0)/(4\pi )\int\limits^(2\pi r)_0 {(i)/(r^2)} \, ds\\\\ = (\mu_0)/(4\pi ){(i * 2\pi r)/(r^2)} \, ds`

Simplify:

`B = (\mu_0 i_(circle))/(2r)\\`

Now, for a straight conductor (assuming of infinite length), we know the magnetic field strength equation to be:

`B = (\mu_0 i_(rod))/(2\pi r)`

Set the two equal.

`(\mu_0 i_(circle))/(2r)= (\mu_0 i_(rod))/(2\pi r)`

Cancel out '2', μ₀, and 'r':

`i_(circle)}= ( i_(rod))/(\pi )`

Plug in the given values and solve.

`40\pi = i_(rod)\\\\i_(rod) = \boxed{125.663 A}`