A long solenoid (1000 turns/m) carries a current of 25 mA and has an inside radius of 2.0 cm. A long wire which is parallel to and 4.0 cm from the axis of the solenoid carries a current of 6.0 A. What is the magnitude of the magnetic field at a point on the axis of the solenoid?

Question

Quizplus · Accepted Answer

Using the formula for magnetic field at the center of a solenoid, 
$$B = \mu_0nI$$ where $n$ is the number of turns per unit length and $I$ is the current. This gives us 
$$B = \mu_0nI = \mu_0\left(\frac{1000 ~	ext{turns}}{	ext{m}}ight)(0.025~	ext{A}) = 1.25664 	imes 10^{-4}~	ext{T}.$$ 
Since the wire is parallel to the axis of the solenoid, it will also produce a magnetic field at the center of the solenoid. Using the formula for magnetic field due to a long straight wire, 
$$B = \frac{\mu_0I}{2\pi r}$$ where $I$ is the current and $r$ is the distance from the wire, we get 
$$B = \frac{\mu_0(6~	ext{A})}{2\pi(0.04~	ext{m})} = 2.3894 	imes 10^{-6}~	ext{T}.$$ 
The net magnetic field at the center of the solenoid is the sum of the magnetic field due to the solenoid and the magnetic field due to the wire, so 
$$B_{net} = B_{solenoid} + B_{wire} = 1.25664 	imes 10^{-4} + 2.3894 	imes 10^{-6} = \boxed{43~\mu	ext{T}}.$$

A Long Solenoid (1000 Turns/m) Carries a Current of 25