PLEASE answer it in the form it is asked above!!!! F = ( __ i + __ j + __ k ) N

Question

PLEASE answer it in the form it is asked above!!!! F = ( __ i + __ j + __ k ) N

3. -/20 pointsWalker4 22.P.044 Each of the 10 turns of wire in a vertical, rectangular loop carries a current of 0.26 A. The loop has a height of 8.0 cm and a width of 15 cm. A horizontal magnetic field of magnitude 0.060 T is oriented at an angle of = 65° relative to the plane of the loop, as indicated in the figure below B/#265 8.0 cm Normal to loop (a) Find the magnetic force on each side of the loop Ftop Fbottom = ( Fi k) N k) N k) N left =( Fright =( (b) Find the net magnetic force on the loop (c) Find the magnetic torque on the loop k) N m (d) If the loop can rotate about a vertical axis with only a small amount of friction, will it end up with an orientation given by = 0, = 90°, or = 180°? -Select- Explain

Explanation / Answer

F = N*I*(LxB)

B = B*cos65i + B*sin65 k = 0.06*cos65i + 0.06*sin65 k

B = 0.025i + 0.054 k

a)

L = 0.15 k

F = 10*0.26*(0.15k x 0.025i + 0.054 k )

Ftop = 10*0.26*0.15 *0.025 = 0i + 0.00975j N + 0 k N

b)

L = -0.15 k

F = 10*0.26*(-0.15k x 0.025i + 0.054 k )

Fbottom = -10*0.26*0.15 *0.025 = 0i - 0.00975j N + 0 k N

(c)

L = -0.08 j

F = 10*0.26*(-0.0.08j x (0.025i + 0.054 k ) )

F = 10*0.26*0.08 *0.025 k - 10*0.26*0.08 *0.054 i

Fleft = -0.11232 i + 0j N + 0.0052 k N

d)

L = 0.08 j

F = 10*0.26*(0.0.08j x (0.025i + 0.054 k ) )

F = 10*0.26*0.08 *0.025 (-k) + 10*0.26*0.08 *0.054 i

Fright = 0.11232 i + 0j N - 0.0052 k N

________________________

Fnet = Ftop + Fbottom + Fleft + Fright = 0

________________

torque = uxB

u = N*I*A = 10*0.26*0.15*0.08 i = 0.0312 i

torque = 0.0312i x (0.025i + 0.054 k)

torque = 0.312*0.054 -j = - 0.016848 j Nm

________

theta = 0

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