The triangular loop of wire shown in the figure carries a current I= 5.00A in th

Question

The triangular loop of wire shown in the figure carries a current I= 5.00A in the direction shown. The loop is in a uniform magnetic field that has magnitude B= 3.00T and the same direction as the current in side PQ of the loop. (Figure 1)

a.) Find the magnitude of the force exerted by the magnetic field on side PQ of the triangle.

b.) Find the magnitude of the force exerted by the magnetic field on side RQ of the triangle.

c.) Find the magnitude of the force exerted by the magnetic field on side PR of the triangle.

f.) What is the magnitude of the net force on the loop.

Explanation / Answer

For direction of the force we will use left hand rule as below,

a)

Angle between B and I , q = 0 deg

Use equation ,

F1= IL x B = ILBsinq = 5.0*0.6*3.0*sin0 = 0 N

b)

angle f in figure = tan^-1(0.6/0.8) =36.87 deg

I = Ix i^ - Iyj^ = Icosf i^ - Isinfj^ = 5.0cos36.87 i^ - 5.0sin36.87j^ = 4.0 i^ - 3j^

L= sqrt(0.6^2+0.8^2) = 1.0

Use equation,

B= B j^

F2= IL x B = 1.0 (4.0 i^ - 3j^) x (3.0 j^) = 12 k^ ………………… directed out of the page

c) Angle between B and I , q = 90 deg

Use equation ,

F3= IL x B = ILBsinq = 5.0*0.8*3.0*sin90 = 12 N –k^ ………………..directed into the page

d) Fnet = F1+F2+F3= 0 + 12 k^ + 12 – k^   = 0 N

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