A long wire carrying 4.5A of current makes two 90o bends, as shown in the figure

Question

A long wire carrying 4.5A of current makes two 90o bends, as shown in the figure (Figure 1) . The bent part of the wire passes through a uniform0.548T magnetic field directed as shown in the figure and confined to a limited region of space.

Part A

Find the magnitude of the force that the magnetic field exerts on the wire.

Part B

Find the direction of the force that the magnetic field exerts on the wire.

Give your answer as an clockwise angle from the rightward direction to the direction of the force.

Explanation / Answer

There are two portions of the wire 1. horizontal, 2. vertical

Let us take horizontal as x-axis, vertical as y-axis and towards us, perpendicular to plane of paper as z-axis

current in horizontal portion = 4.5 i

current in vertical portion = 4.5 j

Magnetic field = 0.548 k

Force acting on a current carrying conductor = L*(i X B)

where L, i and B are length of wire, current and magnetic field respectively.

Force on horizontal portion = 0.6* (4.5 i X 0.548 k) = -1.48 j N

Force on vertical portion = 0.3 * (4.5 j X 0.548k) = 0.74 i N

Total force = 0.74 i -1.48 j N

(a) Magnitude = sqrt(0.74^2 + 1.48^2) = 1.65 N

(b) direction below horizontal, tan(theta) = 1.48/0.74

theta = 63.43 degrees clockwise from rightward direction

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