Chapter 19, Problem 71QAP

To determine

(a)

The current that is needed to generate a magnetic field that has a magnitude of 0.50 T

Answer to Problem 71QAP

The current that is needed to generate a magnetic field of 0.50 T= 3.0 x 10² A(rounded to two significant figures)

Explanation of Solution

Given:

A coil of wire has a diameter of 15 cm = 0.15 m is consisting 250 windings. The magnitude of the magnetic field generated by the coil due to a current that flows through is 0.050 T at a perpendicular distance of 30 cm from the center of the coil.

Formula used:

The z component of the magnetic field(B_z) for a coil of wire with N windings carrying a current i is given by the following equation.

  $\begin{array}{l}{\text{B}}_Z=\frac{{\mu }_{0}Ni{R}^2}{2{(R^2+z^2)}^{3/2}}\to (a)\\ {\text{where B}}_z=\text{ z component of the magnetic field }\\ \text{N = number of windings}\\ \text{R= radius of the loop}\\ \text{z = the perpendicular distance measured from the center of the coil}\\ \text{i = current in the coil }\\ \end{array}$   ${\mu }_0=\text{ permeability of free space}$

Calculation:

It is given that the magnitude of the magnetic field at perpendicular distance 3.0 cm from the center of the coil is 0.50 T. The diameter of the coil is given as 15 cm which implies that the radius of the coil is 7.5 cm. Also, the number of the windings of the coil is given as 250.Substituing these values to equation (a) one could easily find the current that generate the desired magnetic field of 0.50 T.

  $\begin{array}{l}{\mu }_0\text{ = 4}\pi \times {\text{10}}^{-7}\frac{T.m}{A}\text{ ; R=7}\text{.5 cm =0}\text{.075 m;z =3 cm=0}\text{.03m}\\ {\text{N =250; B}}_z=0.50\text{ T}\\ \text{Rearrnging (a);}\\ \text{i =}\frac{B_z\times 2{(R^2+z^2)}^{3/2}}{{\mu }_0.N.R^2}\to (b)\\ \text{substituing to above equation;}\\ \text{i = }{\frac{(0.50\text{ T)}\times \text{2}\left[{(0.075\text{ m)}}^2\text{ +(0}{\text{.030 m)}}^2\right]}{\left(\text{4}\pi \times {\text{10}}^{-7}\frac{T.m}{A}\right)\times 250\times {(0.075\text{ m)}}^2}}^{3/2}\\ \Rightarrow i=2.98\times {10}^2\text{ A=3}\text{.0}\times {\text{10}}^2\text{ A}\end{array}$

Conclusion:

The current that is needed to generate a magnetic field of 0.50 T= 3.0 x 10² A (rounded to two significant figures)

To determine

(b)

The magnitude of magnetic field at the center of the coil at the forehead

Answer to Problem 71QAP

The magnitude of the magnetic field at the center of the forehead= 0.62 T

Explanation of Solution

Given:

A coil of wire has a diameter of 15 cm = 0.15 m is consisting 250 windings. The magnitude of the magnetic field generated by the coil due to a current that flows through is 0.050 T at a perpendicular distance of 30 cm from the center of the coil.

Formula used:

The z component of the magnetic field(B_z) for a coil of wire with N windings carrying a current i is given by the following equation.

  $\begin{array}{l}{\text{B}}_Z=\frac{{\mu }_{0}Ni{R}^2}{2{(R^2+z^2)}^{3/2}}\to (a)\\ {\text{where B}}_z=\text{ z component of the magnetic field }\\ \text{N = number of windings}\\ \text{R= radius of the loop}\\ \text{z = the perpendicular distance measured from the center of the coil}\\ \text{i = current in the coil }\\ \end{array}$

  ${\mu }_0=\text{ permeability of free space}$

Calculation:

It is deduced from part a) that the current that generate a magnetic field of magnitude 0.50 T at a perpendicular distance of 3.0 cm from the center of the coil is 298 A. Now we are asked to calculate the magnitude of the magnetic field at the center of the coil. The diameter of the coil is given as 15 cm which implies that the radius of the coil is 7.5 cm. Also, the number of the windings of the coil is given as 250.Substituing these values to equation (a) one could easily find the magnitude of the magnetic field at the center. Note here that since we have to calculate the magnetic field at the center of the coil the perpendicular distance measured from the center of the coil(z) is 0 cm.

  $\begin{array}{l}{\mu }_0\text{ = 4}\pi \times {\text{10}}^{-7}\frac{T.m}{A}\text{ ; R=7}\text{.5 cm =0}\text{.075 m;z =0 cm=0 m}\\ \text{N =250 i = 298 A;}\end{array}$

Substituting to equation(a);

  $\begin{array}{l}{\text{B}}_z=\frac{\left(\text{4}\pi \times {\text{10}}^{-7}\frac{T.m}{A}\right)\times 250\times 298\text{ A}\times {(0.075\text{ m)}}^2}{\text{2}{\left[{(0.075\text{ m)}}^2\text{ +(0}{\text{.0 m)}}^2\right]}^{3/2}}\\ \Rightarrow {\text{B}}_z=\frac{\left(\text{4}\pi \times {\text{10}}^{-7}\frac{T.m}{A}\right)\times 250\times 298\text{ A}}{2\times 0.075\text{ m}}\text{ = 0}\text{.62 T}\\ \end{array}$

Conclusion:

The magnitude of the magnetic field at the center of the coil = 0.62 T

To determine

(c)

If the current needed in part a) is too high how could one easily achieve the same magnetic field of 0.50 T

Answer to Problem 71QAP

The easiest way that one could compensate for higher currents inside a loop in order to generate a particular magnetic field magnitude is to increase the number of windings in the loop.

Explanation of Solution

Given:

From the calculations in part a) it has been deduced that a current of 298 A is required to generate a magnetic field of 0.50 T from a loop that has 250 windings.

Calculation:

The z component of the magnetic field(B_z) for a coil of wire with N windings carrying a current i is given by the following equation.

  $\begin{array}{l}{\text{B}}_Z=\frac{{\mu }_{0}Ni{R}^2}{2{(R^2+z^2)}^{3/2}}\to (a)\\ {\text{where B}}_z=\text{ z component of the magnetic field }\\ \text{N = number of windings}\\ \text{R= radius of the loop}\\ \text{z = the perpendicular distance measured from the center of the coil}\\ \text{i = current in the coil }\\ \end{array}$

Careful inspection of equation (a) reveals us that B_z is directly proportional to N.So instead of increasing the current(i) one could increase N to achieve higher B_z values.

Conclusion:

The easiest way that one could compensate for higher currents inside a loop in order to generate a particular magnetic field magnitude is to increase the number of windings in the loop