A rectangular 11.0 c m by 18.0 c m circuit carrying an 8.00 A current is oriente

Question

A rectangular 11.0cm by 18.0cm circuit carrying an 8.00A current is oriented with its plane parallel to a uniform 0.720T magnetic field (see the figure (Figure 1) ).

Part A Find the magnitude of the magentic force on each segment of this circuit

Fab,Fbc,Fcd,Fda

Part B Find the direction of the force on the segment Ab

Part C Find the direction of the force on the segment Bc,

Part D Find the direction of the force on the segment cd

Part E Find the direction of the force on the segment da

PArt F find the magnitude of the net force on the entire circuit.

Explanation / Answer

Part A)

Apply F=BILsin(angle)

Fab = (.72)(8)(.18)(sin 90) = 1.94 N

Fbc = 0 since sin 0 = 0

Fcd = (.72)(8)(.18)(sin 90) = 1.94 N

Fda = 0 since sin 0 = 0

Part B)

From the right hand rule, the force is into the page

Part C)

No Force. No Direction

Part D)
From the right hand rule, the force is out of the page

Part E)

No Force. No Direction

Part F

Net force = 0 N (the top and bottom cancel each other out since they are in opposite directions.

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