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Question

(B) The magnetic field due to a long straight wire at a distance $r$ is given by $B_{	ext{wire}} = \frac{\mu_0 I}{2 \pi r}$.Given $I = 5.0 \, A$ and

Vedclass · Accepted Answer

$0.6 	imes 10^{-5} \, T$

Vedclass · Answer

(B) The magnetic field due to a long straight wire at a distance $r$ is given by $B_{	ext{wire}} = \frac{\mu_0 I}{2 \pi r}$.Given $I = 5.0 \, A$ and $r = 10.0 \, cm = 0.1 \, m$.$B_{	ext{wire}} = \frac{4 \pi 	imes 10^{-7} 	imes 5.0}{2 \pi 	imes 0.1} = 1.0 	imes 10^{-5} \, T$.The compass needle aligns with the resultant magnetic field. The deflection is $60^{\circ}$ north of east. From the geometry of the vector addition,$	an 60^{\circ} = \frac{B_{	ext{wire}}}{B_H}$,where $B_H$ is the horizontal component of the earth's magnetic field.Therefore,$B_H = \frac{B_{	ext{wire}}}{	an 60^{\circ}} = \frac{1.0 	imes 10^{-5}}{\sqrt{3}} \approx 0.577 	imes 10^{-5} \, T \approx 0.6 	imes 10^{-5} \, T$.Thus,option $(B)$ is the correct answer.

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