Chapter 20, Problem 80GP

(a)

To determine

Number of turns needed to run the power supply at maximum power.

Answer to Problem 80GP

Number of turns needed to run the power supply at maximum power are $14.6$

Explanation of Solution

Given:

The magnetic field at the center of the coil is given as $\text{B = }\frac{{\mu }_0\text{NI}}{2\text{r}}$ .

Diameter of a circular coil made from a square copper wire of $1.6\text{ mm}$ on a side is $1.2\text{ m}$ .

At a maximum power output of $4\text{ KW}$ power supply produced is $120\text{}\text{V}$

Formula used:

Number of turns is $\frac{L}{2\pi r}$

Calculation:

Resistance in the copper wire is

  $\begin{array}{l}P=\frac{V^2}{R}\\ R=\frac{V^2}{P}\end{array}$

Length of the copper wire wrapped on a square of sided is

  $\begin{array}{c}R={\rho }_{Cu}\frac{L}{A}\\ ={\rho }_{Cu}\frac{L}{d^2}\\ L=\frac{R{d}^2}{{\rho }_{Cu}}\end{array}$

Thus, number of turns of the copper wire is

  $\begin{array}{c}N=\frac{\left(\frac{R{d}^2}{{\rho }_{Cu}}\right)}{2\pi r}\\ =\frac{V^2{d}^2}{2\pi r{\rho }_{Cu}P}\\ \text{ =}\frac{{(120\text{ V})}^2{(1.6\times {10}^{-3}\text{ m})}^2}{2\pi (0.6\text{ m})(1.68\times {10}^{-7}\Omega \cdot \text{m})(4\times {10}^3\text{ W})}\\ \text{= 14}\text{.6 turns}\end{array}$

Conclusion:

  $14.6$ number of turns are needed to run the power supply at maximum power.

(b)

To determine

The magnetic field strength at the center of the coil.

Answer to Problem 80GP

The magnetic field strength at the center of the coilis $5.1\times {10}^{-4}\text{}\text{T}$

Explanation of Solution

Given:

The magnetic field at the center of the coil is given as $\text{B = }\frac{{\mu }_0\text{NI}}{2\text{r}}$ .

Diameter of a circular coil made from a square copper wire of $1.6\text{ mm}$ on a side is $1.2\text{ m}$ .

At a maximum power output of $4\text{ KW}$ power supply produced is $120\text{}\text{V}$

Formula used:

Power is calculated as $P=IV$ .

Calculation:

The magnetic field is calculated as

  $\begin{array}{c}B=\frac{{\mu }_{0}NI}{2r}\\ =\frac{{\mu }_{0}NP}{2rV}\\ =\frac{(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A})(14.6)(4\times {10}^3\text{}\text{W})}{2(0.6\text{ m})(120\text{}\text{V})}\\ \text{=5}\text{.1}\times {\text{10}}^{-4}\text{ T}\end{array}$

Conclusion:

Thus, the magnetic field strength at the center of the coilis $5.1\times {10}^{-4}\text{}\text{T}$

(c)

To determine

To explain: Effect on the magnetic field, if number of turns increases for the same power supply.

Answer to Problem 80GP

If number of turns increases, so will be the magnetic field in the coil.

Explanation of Solution

Given:

The magnetic field at the center of the coil is given as $\text{B = }\frac{{\mu }_0\text{NI}}{2\text{r}}$ .

Since magnetic field is directly proportional to the number of turns. Thus, if number of turns increase, so will be the magnetic field in the coil. If number of turns increases, this suggests that the length of the copper wire from which circular wire is made is greater. If the length increases, so will be the resistance in the wire and the current will subsequently decreases. To keep the magnetic field constant, dependency of number of turns must be removed.

Power is calculated as $\text{P = IV}$ . The magnetic field is now simplified as

  $\begin{array}{l}B=\frac{{\mu }_{0}NI}{2r}=\frac{{\mu }_{0}NP}{2rV}\\ B=\frac{{\mu }_0}{2rV}\frac{R{d}^2}{2\pi {\rho }_{Cu}}\frac{V}{R}=\frac{{\mu }_0{d}^2}{4\pi r^2{\rho }_{Cu}}=\text{constant}\end{array}$

Conclusion:

Thus, increasing number of turns will lead to increase in the magnetic field of the coil.