A current carrying loop (I=0.6A) is placed in a uniform magnetic field (B=0.5T)

Question

A current carrying loop (I=0.6A) is placed in a uniform magnetic field (B=0.5T) as shown in the figure. Each side of the loop is 80cm long. Determine the magnitude and direction of the magnetic force on each side of the loop.

Side AB =

Side BC =

Side CD =

Side DE =

Side EA =

Also, determine the magnitude and direction of the net magnetic force on the loop.

A current carrying loop (I = 0.6A) is (B = 0.5T) as shown in the figure. Each side of the Determine the magnitude and direction Determine the magnitude and direction

Explanation / Answer

F = L(IxB)

Side AB = 0.8(0.6 x 0.5) = 0.24N upwards

Side BC = 0.8(0.6 x 0.5)x sin120 = 0.207N upwards

Side CD = 0.8(0.6 x 0.5)x sin60 = 0.207N downwards

Side DE = 0.8(0.6 x 0.5) = 0.24N downwards

Side EA = 0.8(0.6 x 0.5) x sin0 = 0N

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