Finding the Minimum Magnetic Field for Coil Tipping: A Scientific Analysis

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The discussion revolves around calculating the minimum magnetic field required for a square coil to tip over when subjected to a magnetic force. The key point is understanding that the net torque from the coil's weight is not zero, as it competes with the magnetic torque. Participants clarify that the torque from gravity acts on the coil's center of mass, while the magnetic torque is influenced by the current and the magnetic field strength. The correct approach involves equating the torques from gravity and the magnetic force to determine tipping conditions. Ultimately, recognizing the pivot point and the distribution of forces is crucial for solving the problem accurately.
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a problem in magnetics...

i've been stuck here for long....please help
;
a square coil of side A ,carrying a current I ,is placed on a table and a magnetic field B parallel to one of the edges is switched on. the mass of coil is M
now the question says-"find min magnetic field for which the coil will start tipping over"
.
this is how i went
net magnetic force = zero
net torque due to magnetic force= BI(L^2)
net torque due to wieght=0
.
have seen its solution which says
torque due to mag force=torque due to wieght
nd equates Mg(L/2)=BI(L^2)
which goes over my head.
pleasezzzzzzzz helpppppppppppp
 
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Why do you think the net torque due to weight is 0? You've basically solved the problem once you realize there is a torque from the weight of the coil. If the two torques are competing, then while the loop is resting on the table the torque from gravity is either stronger or equal to the magnetic torque on the loop. However, if the torque from the coil is stronger than the torque from gravity the loops tips over.
 


i can't understand it...
there are 2 arms of loop that experience magnetic force...consider each arm - one would have torque due to magnetic force BI(L^2)/2 and a opposite torque due to weight Mg/4 * L/2 (shouldn't mass of only the arm be considered?) and on second one both torques in same direction
the torque due to weight on each arm is in opposite directions and cancels out
 


pleasezzzzzzzzzzzzz replyyyyyyyyyyyy
 


You're viewing the loop as floating. The loop is on the table so only the end that has torque in the direction of free space will move.
 
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and so?
 


So ignore the torque at the other end because that end cannot move.
 


ok then...i get Mg/4 *L\2 = BI(L^2)/2 ...which is still wrong!
 


No, take your pivot point to be the far leg (because this leg will always be pushed into the table by both gravity and the magnetic field it will actually be anchored pretty well), and now the torque due to gravity acts on the center of mass.
 
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kkkk...got it...thanks
 
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