As shown in the figure, a rectangular loop with a length of 20.0 cm and a width

Question

As shown in the figure, a rectangular loop with a length   of 20.0 cm and a width w of 15.0 cm has 20 turns and carries a current of 0.21 A counterclockwise around the loop when viewed from the positive x-axis. A horizontal (parallel to the x-z plane) magnetic field of magnitude 0.057 T is oriented at an angle of 65° relative to the perpendicular to the loop (the positive x-axis). (Assume the length and width are measured along the z and y-axes, respectively.)

(a) Find the components of the magnetic force acting on each side of the loop.

Components of the magnetic force acting on the top section of the loop.

F top x = N

F top y= N

F top z = N

Components of the magnetic force acting on the bottom section of the loop.

F bottom x= N

F bottom y= N

F bottom z= N

Components of the magnetic force acting on the left section of the loop.

F left x= N

F left y= N

F left z= N

Components of the magnetic force acting on the right section of the loop.

F right x= N

F right y = N

F right z= N

(b) Find the net magnetic force on the loop. (Enter the magnitude only.)

Fnet= N

(c) Find the magnetic torque on the loop.

magnitude = ? N*M

Explanation / Answer

let i,j and k are unit vectors along +ve x, +ve y and +ve z direction

magnetic field in vector notation:

B=0.057*(cos(65) i +sin(65) k)

=0.0241 i + 0.0517 k

magnetic force=number of turns*current*cross product of length and magnetic field

part a:

for top section of the loop,

current is along +ve z axis.

length=20 cm=0.2 m

then force=20*0.21*cross product of (0.2 k and (0.0241 i + 0.0517 k))

=20*0.21*0.2*0.0241 j

=0.020244 j

so Ftop x=0

Ftop y=0.020244 N

Ftop z=0

part ii:

for bottom section, current in along -ve z axis.

length is 0.2 m

Fbottom x=0

Fbottom y=-0.020244 N

Fbottom z=0

part iii.

for left section, current is along -ve y axis.

length is 0.15 m

force=20*0.21*cross product of (-0.15 j and B)

Fleft x=-0.032546 N

Flft y=0

Fleft z=0.015176 N

part iv.

for right section, current is along +ve y axis

length is 0.15 m

Fright x=0.032546 N

Fright y=0

Fright z=-0.015176 N

pat b:

net magnetic force=force on left section + force on right section + force on top section + force on bottom section

=0

part c:

magnetic torque=cross product of m and B

where m is magnetic moment=number of turns*current*area*unit vector along the area

=20*0.21*0.2*0.15 i

=0.126 i

then torque=cross product of 0.0063 i and B

=-0.00651 j

so magnitude of magnetic torque is 0.00651 N.m

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