1)A rectangular 11.0cm by 18.0cm circuit carrying an 8.00A current is oriented with its plane parallel to a uniform 0.700T magnetic field (see the figure (Figure 1) ).
figure--> http://s16.postimg.org/olp71b0kh/123456.jpg
A)Find the magnitude of the magnetic force on each segment ( Fab,Fbc,Fcd,Fda) of this circuit.
B)Find the magnitude of the net force on the entire circuit.
3 Answers
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A
Fad and Fbc the force is zero ( parallel to magnetic lines of flux)
Fab = - Fcd = currentlength of sidemagnetic field strength
magnitude of Fab and Fcd = 80.180.7 = 100.8 N
B
net force = 100.8 + 100.8 = 201.6 N
PS
torque about axis through middle of ab, cd = currentareamagnetic field strength = 80.110.18*0.7
= 0.0678 Nm
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F_ab= 1.1N
F_bc= 0N
F_cd= 1.1N
F_da= 0N
net force = 0N
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In the above best answer, mastering physics accepts the answer of part a as 1.0 N. The answer user Anthony provided is 100x larger than the answer it accepts.
for part b, the net force is 0.