Chapter 19, Problem 70QAP

To determine

(a)

The magnitude of magnetic field at y =0 cm according to the adapted coordinate system

Answer to Problem 70QAP

The magnitude of the magnetic field at (y =0 cm) = 0 T

Or in other words there is no magnetic field at y= 0 cm

Explanation of Solution

Given:

Two long straight wires are positioned such that they are parallel to x at y=2.5 cm and y= -2.5 cm as shown in the diagram below. Each wire carries a current of 16 A.

Formula used:

  $\begin{array}{l}B=\frac{{\mu }_{0}i}{2\pi r}\to (a)\\ \text{where B= Magnitude of the magentic field due to a long, straight wire}\\ {\mu }_0=\text{ permeability of free space}\\ \text{r = distance from the wire to the location where the field is measured}\end{array}$

Right hand rule for the field directionIf you point your right thumb in the direction of the current and curl your fingers, the magnetic field curls around the field lines in the direction of the curled fingers of your right hand.

Calculation:

One can make use of the superposition to calculate the magnitude of the magnetic field at designated location of the y axis.

Let us denote the magnitude of the magnetic field generated by the top wire as B_topand the magnetic field generated by the wire below as B_bottom.

Applying the right hand rule one could clearly see that B_top acts in the -z direction while B_bottomacts in the +z direction.

Let B= total magnitude of the magnetic field at y=0 cm

Then B = B_top + B_bottom

Substituting for equation (a) we could find B as follows where r_top= distance from the top wire to y=0 and r_bottom= distance from the bottom wire to y=0.

  $\begin{array}{l}\text{B = }\left(\frac{-{\mu }_{0}i}{2\pi r_{top}}\right)\text{ + }\left(\frac{{\mu }_{0}i}{2\pi r_{bottom}}\right)\\ B\text{ =}\left(\frac{{\mu }_{0}i}{2\pi }\right)\left\{\frac{-1}{r_{top}}\text{ + }\frac{1}{r_{bottom}}\right\}\\ r_{top}=r_{bottom}=2.5\text{ cm = 0}\text{.025m ;}\\ B=\left(\frac{{\mu }_{0}i}{2\pi }\right)\left\{\frac{-1}{0.025\text{ m}}\text{ + }\frac{1}{0.025\text{ m}}\right\}\\ \Rightarrow B=0T\end{array}$

Conclusion:

The magnitude of the magnetic field at (y =0 cm) = 0 T

Or in other words there is no magnetic field at y= 0 cm

To determine

(b)

The magnitude of magnetic field at y = 1.0 cm according to the adapted coordinate system

Answer to Problem 70QAP

The magnitude of the magnetic field at (y =1.0 cm) = (-1.2 x 10-4)T

Explanation of Solution

Given:

Two long straight wires are positioned such that they are parallel to x at y=2.5 cm and y= -2.5 cm as shown in the diagram below. Each wire carries a current of 16 A.

Calculation:

One can make use of the superposition to calculate the magnitude of the magnetic field at designated location of the y axis.

Let us denote the magnitude of the magnetic field generated by the top wire as B_topand the magnetic field generated by the wire below as B_bottom.

Applying the right handrule, one could clearly see that B_top acts in the -z direction while B_bottomacts in the +z direction.

Let B= total magnitude of the magnetic field at y=1.0 cm

Then B = B_top + B_bottom

Substituting for equation (a) we could find B as follows where r_top= distance from the topwire to y=1.0cm and r_bottom= distance from the bottom wire to y=1.0 cm.

  $\begin{array}{l}\text{B = }\left(\frac{-{\mu }_{0}i}{2\pi r_{top}}\right)\text{ + }\left(\frac{{\mu }_{0}i}{2\pi r_{bottom}}\right)\\ B\text{ =}\left(\frac{{\mu }_{0}i}{2\pi }\right)\left\{\frac{-1}{r_{top}}\text{ + }\frac{1}{r_{bottom}}\right\}\\ r_{top}=1.5\text{ cm = 0}{\text{.015 m and r}}_{bottom}=3.5\text{ cm = 0}\text{.035 m}\\ B=\left(\frac{{\mu }_{0}i}{2\pi }\right)\left\{\frac{-1}{0.015\text{ m}}\text{ + }\frac{1}{0.035\text{ m}}\right\}\\ B\text{ =}\left(\frac{(4\pi \times {10}^{-7}\frac{T.m}{A})\times (16A)}{2\pi }\right)\left\{\frac{-1}{0.015\text{ m}}\text{ + }\frac{1}{0.035\text{ m}}\right\}\\ \Rightarrow B\text{ = (-1}\text{.2 }\times {\text{10}}^{-4}\text{) T }\end{array}$

Conclusion:

The magnitude of the magnetic field at (y =1.0 cm) = (-1.2 x 10-4)T

To determine

(c)

The magnitude of magnetic field at y = 4.0 cm according to the adapted coordinate system

Answer to Problem 70QAP

The magnitude of the magnetic field at (y =4.0 cm) = (2.6 x 10-4)T

Explanation of Solution

Given:

Two long straight wires are positioned such that they are parallel to x at y=2.5 cm and y= -2.5 cm as shown in the diagram below. Each wire carries a current of 16 A.

Calculation:

One can make use of the superposition to calculate the magnitude of the magnetic field at designated location of the y axis.

Let us denote the magnitude of the magnetic field generated by the top wire as B_topand the magnetic field generated by the wire below as B_bottom.

Applying the right hand rule, one could clearly see that B_top acts in the +z direction while B_bottomalso acts in the +z direction.

Let B= total magnitude of the magnetic field at y=4.0 cm

Then B = B_top + B_bottom

Substituting for equation (a) we could find B as follows where r_top= distance from the top wire to y=4.0cm and r_bottom= distance from the bottom wire to y=4.0 cm.

  $\begin{array}{l}\text{B = }\left(\frac{-{\mu }_{0}i}{2\pi r_{top}}\right)\text{ + }\left(\frac{{\mu }_{0}i}{2\pi r_{bottom}}\right)\\ B\text{ =}\left(\frac{{\mu }_{0}i}{2\pi }\right)\left\{\frac{-1}{r_{top}}\text{ + }\frac{1}{r_{bottom}}\right\}\\ r_{top}=1.5\text{ cm = 0}{\text{.015 m and r}}_{bottom}=6.5\text{ cm = 0}\text{.065 m}\\ B=\left(\frac{{\mu }_{0}i}{2\pi }\right)\left\{\frac{1}{0.015\text{ m}}\text{ + }\frac{1}{0.065\text{ m}}\right\}\\ B\text{ =}\left(\frac{(4\pi \times {10}^{-7}\frac{T.m}{A})\times (16A)}{2\pi }\right)\left\{\frac{1}{0.015\text{ m}}\text{ + }\frac{1}{0.065\text{ m}}\right\}\\ \Rightarrow B\text{ = (2}\text{.6 }\times {\text{10}}^{-4}\text{) T }\end{array}$

Conclusion:

The magnitude of the magnetic field at (y =4.0 cm) = (2.6 x 10-4)T