A wire 70.0 mm long is bent in a right angle such that the wire starts at the or

Question

A wire 70.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 25.0 mm , y = 0, and then in another straight line from x = 25.0 mm , y = 0 to x = 25.0 mm , y = 45.0 mm . The wire is in an external uniform 0.500-T magnetic field in the +z direction, and the current through the wire is 3.30 A , directed from the origin into the wire.

Part A Determine the magnitude of the magnetic force exerted by the external field on the wire.

Part B

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

Part C

The wire is removed and replaced by one that runs directly from the origin to the location x = 25.0 mm , y = 45.0 mm . If the current through this wire is also 3.30 A , what is the magnitude of the magnetic force exerted by the external field on the wire?

Part D

The wire is removed and replaced by one that runs directly from the origin to the location x = 25.0 mm , y = 45.0 mm . If the current through this wire is also 3.30 A , what is the magnitude of the direction force exerted by the external field on the wire measured clockwise from the positive x axis.?

Explanation / Answer

(a)

The separation between the end points is

l = sqrt (25 mm)^2 + (45 mm)^2 = 51.47 mm

force is

F = B i l = 0.5 ( 3.3) 51.47 * 10^-3 m = 84.93 * 10^-3 N

(b)

angle

tan theta = 45/ 25

theta = tan^-1 ( 45/ 25) = 60.94

the angle that the force makes is

theta' = theta - 90 =60.94 - 90= -29.05 degree or

330.94 degree with positive x axis in counterclock wise direction

(c)

F = B i l = 0.5 ( 3.3) 51.47 * 10^-3 m = 84.93 * 10^-3 N

(d)

it would have same direction 330.94

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.