Finding Magnitude & Direction of Force & Torque Exerted by B on Coil

AI Thread Summary
The discussion focuses on calculating the force and torque exerted by a magnetic field on a coil of wire with specific dimensions and current. The magnetic field is defined in both x and z directions, complicating the calculation of force. Participants emphasize the use of the vector equation for force, F = I L × B, and the importance of performing the cross product correctly. One contributor calculates the force magnitude but is reminded that the components must be treated as a vector rather than simply summed. The conversation highlights the need for careful vector analysis in magnetic field problems.
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Homework Statement



A coil of wire consisting of 40 rectangular loops, with width 16 cm and height 30 cm, is placed in a constant magnetic field given by
B = 0.055Tx + 0.210T.z
The coil is hinged to a fixed thin rod along the y-axis (along segment da in the figure) and is originally located in the xy-plane. A current of 0.150 A runs through the wire.http://www.webassign.net/bauerphys1/27-p-051.gifWhat are the magnitude and the direction of force, Fbc, that B exerts on segment bc of the coil?
magnitude Fbc =
direction °
What are the magnitude and the direction of the torque, t, that B exerts on the coil?
magnitude t =

The Attempt at a Solution



i am just lost

i have the formulas but I'm not sure how to find the magnitube for B
hints pleasE?
 
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What's the formula for the force of a magnetic field on a current carrying wire of a given length?
 


F = ILB
yeah but B is in x and z direction so do i just do

0.15 A * 0.3 M * (0.055T x + 0.210T z)

and then do i square x and z then rad it? i tried that but it's not right
 


yjk91 said:
F = ILB
yeah but B is in x and z direction so do i just do

0.15 A * 0.3 M * (0.055T x + 0.210T z)

and then do i square x and z then rad it? i tried that but it's not right

It's a vector equation:
\vec{F} = I \; \vec{L} \times \vec{B}
In this case you have 40 wires of length 30cm all carrying the same current I, so multiply the result by 40.
\vec{F} = 40I \; \vec{L} \times \vec{B}
You should be able to write vectors for both L and B and perform the cross product (do it manually, it's probably easier). The result will be your force vector.
 


so

40 * .15 * 0.3 X (0.055x + 0.210z)


using the cross product i get (0.3 * .21, 0 , -.3*.055)
if i add them i get 0.0465

and 40 * .15 * .0465 = .279 N

is this right?
 


yjk91 said:
so

40 * .15 * 0.3 X (0.055x + 0.210z)


using the cross product i get (0.3 * .21, 0 , -.3*.055)
if i add them i get 0.0465

and 40 * .15 * .0465 = .279 N

is this right?

The cross product yields a vector, so you can't just add the components. You have to take the magnitude of the vector to find the magnitude of the force. First write out all three components of the force vector.
 
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