Question

29. What are the magnetic field strength and direction at points 1, 2, and 3 in the figure shown? 1. 2.0 cm 10 A 2. 4.0 cm 10 A 2.0 cm 3-

Answer 1

i) Point 1
Magnetic field due to the top wire,
Bt = (_$\mu$o * I) / (2$\pi$ * r), out of the page
Where _{$\mu$ o} is the vacuum permeability, I is the current, and r = 2 cm.

Magnetic field due to the bottom wire,
Bb = (_$\mu$o * I) / (2$\pi$ * 3r), into the page

Total magnetic field = Bt + Bb
= (_$\mu$o * I) / (2$\pi$ * r) - (_$\mu$o * I) / (2$\pi$ * 3r) [out of the page direction is taken as positive]
= (_$\mu$o * I) / (2$\pi$ * r) [1 - 1/3]
Substituting values,
B = [4$\pi$ * 10^-7 * 10] / [2$\pi$ * 2 * 10^-2] * 0.67
= 6.67 * 10^-5 T, out of the page.
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ii) Point 2
Magnetic field due to the top wire,
Bt = (_$\mu$o * I) / (2$\pi$ * r), into the page

Magnetic field due to the bottom wire,
Bb = (_$\mu$o * I) / (2$\pi$ * r), into the page

Total magnetic field = Bt + Bb
= (_$\mu$o * I) / (2$\pi$ * r) + (_$\mu$o * I) / (2$\pi$ * r)
= (_$\mu$o * I) / (2$\pi$ * r) * 2
Substituting values,
B = [4$\pi$ * 10^-7 * 10] / [2$\pi$ * 2 * 10^-2] * 2
= 20 * 10^-5 T, into the page.
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iii) Point 3
Magnetic field due to the top wire,
Bt = (_$\mu$o * I) / (2$\pi$ * 3r), into the page

Magnetic field due to the bottom wire,
Bb = (_$\mu$o * I) / (2$\pi$ * r), out of the page

Total magnetic field = Bt + Bb
= (_$\mu$o * I) / (2$\pi$ * r) - (_$\mu$o * I) / (2$\pi$ * 3r) [out of the page direction is taken as positive]
= (_$\mu$o * I) / (2$\pi$ * r) [1 - 1/3]
Substituting values,
B = [4$\pi$ * 10^-7 * 10] / [2$\pi$ * 2 * 10^-2] * 0.67
= 6.67 * 10^-5 T, out of the page.