A long wire carrying I = 6.75A of current makes two bends, as shown in the figur

Question

A long wire carrying I = 6.75A of current makes two bends, as shown in the figure (Figure 1) . The bent part of the wire passes through a uniform 0.270T magnetic field directed as shown in the figure and confined to a limited region of space.

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

F= ________ N

B) Find the direction of the force that the magnetic field exerts on the wire. Give your answer as a clockwise angle from the left-hand straight segment.

? = _______ degrees

Explanation / Answer

There are two segments that are horizontal. The diagonal portion has a horizontal length found by the tangent function

tan(30) = 40/x

x = 69.3 cm

Since the total width is 1 m, the two horizontal pieces have a length of 1 - .693 = .307 m

Apply F = BIL

F = (.27)(6.75)(.307) = .559 N directed upwards (North)

For the diagonal portion, its length is found using the sin function

sin (30) = 40/x

x = 80 cm

Apply F = BIL

F = (.27)(6.75)(.8) = 1.458 directed at 60 degrees North of West

We need the North and West components

F(North) = 1.458(sin 60) = 1.262 N

F(West) = 1.458(cos 60) = 0.729 N

The net north = 1.262 + 0.559 = 1.821 N

The net west = 0.729 N

Now we can find the total using the pythagorean theorem

Net = sqrt[(1.821)^2 + (0.729)^2]

Net = 1.961 N

The direction is found using the tangent function

tan (angle) = 1.821/0.729

angle = 68.18 degrees North of West

In summary, the force is 1.961 N 68.2 degrees clockwise from the left straight segment.

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