A long straight wire carries a current of 40 A in a region where a uniform external magnetic field has a 30-μT magnitude and is parallel to the current. What is the magnitude of the resultant magnetic field at a point that is 20 cm from the wire?
A) 70 μT
B) 40 μT
C) 10 μT
D) 50 μT
E) 36 μT

Question

Quizplus · Accepted Answer

The magnetic field produced by the wire is given by
$B_{wire} = \frac{\mu_0}{2\pi}\frac{I}{r}$,
where $r$ is the distance from the wire and $\mu_0$ is the magnetic constant. Plugging in the given values, we get
$B_{wire} = \frac{(4\pi 	imes 10^{-7} 	ext{T}\cdot	ext{m}/	ext{A})(40 	ext{ A})}{2\pi (0.2 	ext{ m})} = 1 	ext{ mT}$.
The external magnetic field also contributes to the total magnetic field at the point, and since it is parallel to the wire, it simply adds to the field produced by the wire. Therefore, the total magnetic field at the point is
$B_{total} = B_{wire} + B_{external} = (1 	ext{ mT}) + (30 	ext{ }\mu	ext{T}) = 1030	ext{ }\mu	ext{T}$.
Converting to Tesla and rounding to two significant figures, we get $B_{total}= 1.0 	ext{ mT} = \boxed{	ext{(D) }50	ext{ }\mu	ext{T}}$.

A Long Straight Wire Carries a Current of 40 a in a Region