A long solenoid (diameter = 5.00 cm)is wound with 2000 turns per meter of thin wire through which a constant current is maintained.A wire carrying 8.50 A is inserted along the axis of the solenoid.The magnitude of the magnetic field at a point 5.00 mm from the axis is 0.650 mT.What is the current through the solenoid?

Question

Quizplus · Accepted Answer

The magnetic field inside a solenoid is given by $B = \mu_0 n I$, where $B$ is the magnetic field, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{Tm/A}$), $n$ is the number of turns per unit length, and $I$ is the current. The magnetic field due to the wire inside the solenoid adds to the field of the solenoid itself. However, at a point 5.00 mm from the axis, the contribution from the wire is negligible, and the field is primarily from the solenoid. Given $B = 0.650 \, \text{mT}$ and $n = 2000 \, \text{turns/m}$, we solve for $I$. Rearranging the formula gives $I = B / (\mu_0 n)$. Substituting the given values, $I = 0.650 \times 10^{-3} / (4\pi \times 10^{-7} \times 2000)$, which calculates to approximately 0.220 A.

A Long Solenoid (Diameter = 5