Chapter 19, Problem 83QAP

To determine

(a)

Magnitude and direction of the magnetic field

Answer to Problem 83QAP

Net magnetic field $=2.64*{10}^{-6}T$ in west direction

Explanation of Solution

Given info:

Current = $100$ A

Distance between lines = $3.00$ m

Distance to magnetic field point = $15.0$ m

Formula used:

  $B=\frac{{\mu }_{0}I}{2\pi r}$

  $I=$ Current

  ${\mu }_0=$ Magnetic constant

  $r=$ Radial distance

  $B=$ Magnetic field

Calculation:

  

1. Net magnetic field,

  $B_{net}=B_1\sin \theta +B_2\sin \theta$

But $B_1=B_2$

  $B_{net}=2B_1\sin \theta =\frac{2{\mu }_{0}I}{2\pi r}\sin \theta =\frac{2*4\pi *{10}^{-7}N{A}^{-2}*100A}{2\pi *\sqrt{{15}^2+{1.5}^2}m}*\frac{15m}{\sqrt{{15}^2+{1.5}^2}m}=2.64*{10}^{-6}T$

Conclusion:

Net magnetic field $=2.64*{10}^{-6}T$ in west direction

To determine

(b)

Magnitude and direction of the magnetic field

Answer to Problem 83QAP

Net magnetic field $=2.64*{10}^{-7}T$ in vertical upward direction

Explanation of Solution

Given info:

Current = $100$ A

Distance between lines = $3.00$ m

Distance to magnetic field point = $15.0$ m

Formula used:

  $B=\frac{{\mu }_{0}I}{2\pi r}$

  $I=$ Current

  ${\mu }_0=$ Magnetic constant

  $r=$ Radial distance

  $B=$ Magnetic field

Calculation:

Direction is west

Net magnetic field,

  $B_{net}=B_1\cos \theta +B_2\cos \theta$

But $B_1=B_2$

  $B_{net}=2B_1\cos \theta =\frac{2{\mu }_{0}I}{2\pi r}\cos \theta =\frac{2*4\pi *{10}^{-7}N{A}^{-2}*100A}{2\pi *\sqrt{{15}^2+{1.5}^2}m}*\frac{1.5m}{\sqrt{{15}^2+{1.5}^2}m}=2.64*{10}^{-7}T$

Conclusion:

Net magnetic field $=2.64*{10}^{-7}T$ in vertical upward direction

To determine

(c)

Magnitude and direction of the magnetic field

Answer to Problem 83QAP

Net magnetic field $=2.64*{10}^{-6}T$ in south direction

Explanation of Solution

Given info:

Current = $100$ A

Distance between lines = $3.00$ m

Distance to magnetic field point = $15.0$ m

Formula used:

  $B=\frac{{\mu }_{0}I}{2\pi r}$

  $I=$ Current

  ${\mu }_0=$ Magnetic constant

  $r=$ Radial distance

  $B=$ Magnetic field

Calculation:

  

Net magnetic field,

  $B_{net}=B_1\sin \theta +B_2\sin \theta$

But $B_1=B_2$

  $B_{net}=2B_1\sin \theta =\frac{2{\mu }_{0}I}{2\pi r}\sin \theta =\frac{2*4\pi *{10}^{-7}N{A}^{-2}*100A}{2\pi *\sqrt{{15}^2+{1.5}^2}m}*\frac{15m}{\sqrt{{15}^2+{1.5}^2}m}=2.64*{10}^{-6}T$

Direction is south

Conclusion:

Net magnetic field $=2.64*{10}^{-6}T$ in south direction

To determine

(d)

Magnitude and direction of the magnetic field

Answer to Problem 83QAP

Yes, could possibly interfere with migration

Explanation of Solution

Given info:

Current = $100$ A

Distance between lines = $3.00$ m

Distance to magnetic field point = $15.0$ m

Formula used:

  $B=\frac{{\mu }_{0}I}{2\pi r}$

  $I=$ Current

  ${\mu }_0=$ Magnetic constant

  $r=$ Radial distance

  $B=$ Magnetic field

Calculation:

  

Yes, could possibly interfere with bird migration

Conclusion:

Yes, could possibly interfere with migration