Question

A closely wound coil has a radius of 6.00 cm and carries a current of 2.50 A. How many turns must it have if, at a point on the coil axis 6.00 cm from the centre of the coil, the magnetic field is$\mathbf{6}\mathbf{.}\mathbf{39}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{T}$.

Short answer

The number of turns is 69.

Step-by-step solution

Concept of the magnetic field

The magnetic field due to a loop of radius a at any point x on the axis of loop is given by:

${\mathrm{B}}_{\mathrm{x}}=\frac{{\mu }_0{\mathrm{Nla}}^2}{2{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}^{3/2}}$

Where N is the number of turns, and I is the current passing through the loop.

Calculate the number of terms

Consider a closely wound coil that has a radius of, and it carries a current ofWe need to find the number of turns such that the magnetic field is BCat a point on the coil axis 6.00 cm from the center of the coil.

To calculate N, we have:

$\mathrm{N}=\frac{2{\mathrm{B}}_{\mathrm{x}}{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}^{3/2}}{{\mu }_0{\mathrm{a}}^2}$

Substitute the values, and we get;

role="math" localid="1668249432744" $\begin{array}{r}\mathrm{N}=\frac{2\left(6.39\times {10}^{-4}\mathrm{T}\right){\left[(0.0600\mathrm{m}{)}^2+(0.0600\mathrm{m}{)}^2\right]}^{3/2}}{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(2.50\mathrm{A}\right)(0.0600\mathrm{m}{)}^2}\\ =69\end{array}$

Therefore, the number of turns is 69.