Question

A long horizontal wire carries 24 A of current due north. What is the net magnetic field 22 cm due west of the wire if the Earth's field there points downward, 42

Answer 1

a)

Magnetic field Due to wire

$$ B=\frac{\mu_{o} I}{2 \pi r}=\frac{\left(4 \pi \times 10^{-7}\right)(24)}{2 \pi \times 0.22}=2.18 \times 10^{-5} T $$

Net Magnetic field is the vector sum of X-component and Y-component of magnetic field \(\mathrm{X}\)-component

$$ \begin{aligned} &B_{x}=B_{\text {earth }} \operatorname{Cos} 42=\left(5 \times 10^{-5}\right) \operatorname{Cos} 42=3.72 \times 10^{-5} \mathrm{~T} \\ &\text { Y-Component } \\ &B_{y}=B_{\text {wire }}-B_{\text {earth }} \operatorname{Sin} 42=\left(2.18 \times 10^{-5}\right)-\left(5 \times 10^{-5}\right) \sin 42 \\ &B_{y}=-1.17 \times 10^{-5} T \end{aligned} $$

Net Magnetic field

$$ B=\sqrt{B_{x}^{2}+B_{y}^{2}}=3.9 \times 10^{-5} T-\text { or }-39 \mu T $$

b)

Direction

$$ \theta=\operatorname{Tan}^{-1}\left(\frac{B_{y}}{B_{x}}\right)=-17.4^{0} $$

\(17.4^{\circ}\) Below the horizontal

Answer 2

magnetic field B due to wire is B = uoi/2pir

B1 = 4pie-7 * 24/(2pi*0.22)

B1 = 2.18e-5 T

B2 = 5 e-5 T

angle between them is 42 deg

so Bnet^2 = B1^2 +B2^2 + 2B1B2 cos 42

Bnet^2 = 2.18^2 + 5^2 + (2* 2.18* 5* cos 42)

Bnet = 6.77 *10^-5 T
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apply angle tan theta = B2/B1

tan theta = 5/2.18 = 2.29

theta = 66.44 deg

and be;ow the horizontal it is 360-66.44 = 329.52 deg