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

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`

Step-by-step explanation:

I = Current

B = Magnetic field

Separation between end points is

`l=√(40^2+55^2)\\\Rightarrow l=68.00735\ mm`

Effective force is given by

`F=IlB\\\Rightarrow F=5.1* 68.00735* 10^(-3)* 0.3\\\Rightarrow F=0.1040512455\ N`

The force is 0.1040512455 N

`tan\theta=(55)/(40)\\\Rightarrow \theta=tan^(-1)(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* √(20^2+35^2)* 10^(-3)* 0.3\\\Rightarrow F=0.05925\ N`

The force is 0.05925 N

`tan\theta=(35)/(20)\\\Rightarrow \theta=tan^(-1)(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`