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A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?
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$1.80 \textrm{ T}$

Solution Video

OpenStax College Physics Solution, Chapter 22, Problem 37 (Problems & Exercises) (0:36)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The force on the current-carrying wire is the current times the wire's length times the magnetic field's strength times sine of the angle between the wire, the current and the magnetic field. So we'll divide by <i>Il</i> and sine theta in order to solve for <i>B</i>. So <i>B</i> is <i>F</i> over <i>Il</i> sine theta. So that's 2.61 Newtons divided by 30 Amps times four times ten to the minus two meters times sine of 90 which is one. And that gives 1.80 Tesla.