A straight 1.0-m long wire is carrying a current.The wire is placed perpendicular to a 0.20-T $\vec { B }$ field.If the magnetic force exerted on the wire has a magnitude of 0.60 N,what is the current in the wire?

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Quizplus · Accepted Answer

The magnetic force on a current-carrying wire is given by $ F = I \cdot L \cdot B \cdot \sin(\theta) $. Here, $ F = 0.60 \, \text{N} $, $ L = 1.0 \, \text{m} $, $ B = 0.20 \, \text{T} $, and $ \theta = 90^\circ $ (since the wire is perpendicular to the field, $\sin(90^\circ) = 1$). Solving for $ I $, we get $ I = \frac{F}{L \cdot B} = \frac{0.60}{1.0 \cdot 0.20} = 3.0 \, \text{A} $.

A Straight 1 B⃗\vec { B }B FieldIf the Magnetic Force Exerted on the Wire Has a Is

A Straight 1 $\vec { B }$ FieldIf the Magnetic Force Exerted on the Wire Has a Is