A long wire carrying 4.50 A of current makes two 90.0 o bends, as shown in the f

Question

A long wire carrying 4.50A of current makes two 90.0o bends, as shown in the figure (Figure 1) . The bent part of the wire passes through a uniform 0.230T magnetic field directed as shown in the figure and confined to a limited region of space.

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

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

A long wire carrying 4.50A of current makes two 90.0o bends, as shown in the figure (Figure 1) . The bent part of the wire passes through a uniform 0.230T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude of the force that the magnetic field exerts on the wire. Find the direction of the force that the magnetic field exerts on the wire.

Explanation / Answer

a) Equation needed: F=(I)(L)(B)*sin(phi) ----angle used for this is 90 so sin(phi)=1

First you need to consider this problem in terms of vectors (x and y components)

given: total "x" length = 0.6m (60/100)

total "y" length = 0.3 (30/100)

1) find the force for the 60 cm section

Fx=(I)(L)(B)

you can consider the 60 cm section as just a flat wire with no y direction

2) find the force for the 30 cm section

Fy=(I)(L)(B)

you can consider the 30 cm section as just a flat wire with no x direction

Now, you have the components of force in the wire, you just need to find out what the combination of these two components is.

In order to find the answer, you need to use the quadratic formula.

(Fx)2+(Fy)2=(F)2

When I computed this all out I got: 0.694 N

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