A rectangular 10.0 cm by 18.0 cm circuit carrying an 8.00 A current is oriented

Question

A rectangular 10.0 cm by 18.0 cm circuit carrying an 8.00 A current is oriented with its plane parallel to a uniform 0.710 T magnetic field (see the figure).

a)Find the magnitude of the magnetic force on each segment (ab, bc, etc)of this circuit.

b)Find the direction of the force on the segment ab.

c)Find the direction of the force on the segment bc.

d)Find the direction of the force on the segment cd.

e)Find the direction of the force on the segment da.

f) Find the magnitude of the net force on the entire circuit.

Explanation / Answer

Part A)

We need to apply the formula F = BILsin for each segment

For ab, F = (.71)(8)(.18)(sin 90) = 1.022 N

For bc, F = (.71)(8)(.1)(sin 0) = 0 N

For cd, F = (.71)(8)(.18)(sin 90) = 1.022 N

For da, F = (.71)(8)(.1)(sin 0) = 0 N

Part B)

By the right hand rule, the direction of the force on ab is into the page (- z direction)

Part C)

Since there is no force on segment bc, there is no direction

Part D)

By the right hand rule, the direction of the force on cd is out of the page (+ z direction)

Part E)

Since there is no force on segment da, there is no direction

Part F)

The net force is zero. Segment ab cancels segment cd. Add the zero's from bc and da, you get a net zero force.

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