Question
A solenoid has 4850 turns per meter and radius $3.3 \mathrm{cm}$. The magnetic field inside has magnitude 0.24 T. What is the current in the solenoid?
Step 1
We know the formula for the magnetic field inside a solenoid: B = Ī¼ā * n * I, where B is the magnetic field, Ī¼ā is the permeability of free space (4Ļ Ć 10ā»ā· Tm/A), n is the number of turns per meter, and I is the current. Show more…
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