Question

A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm , y = 0, and then in another straight line from x = 40.0 mm , y = 0 to x = 40.0 mm , y = 55.0 mm . The wire is in an external uniform 0.300-T magnetic field in the +z direction, and the current through the wire is 5.10 A , directed from the origin into the wire.
A/Determine the magnitude of the magnetic force exerted by the external field on the wire.

B/Determine the direction of the magnetic force exerted by the external field on the wire measured clockwise from the positive x axis.

C/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the magnetic force exerted by the external field on the wire?

D/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the direction force exerted by the external field on the wire measured clockwise from the positive x axis.?

Answer 1

Answer:

0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

0.05925 N

$29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

Explanation:

I = Current

B = Magnetic field

Separation between end points is

$l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm$

Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

The direction is $29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$