freezeice04
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- Apr 29, 2010
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- HSC
- 2012
Dot Point 1.5.2 part (c)
The diagram shows a horizontal 10cm square loop carrying 1.5 A in a magnetic field of strength 0.25T
c) Predict the force on side AC of the loop if it was made into a coil with 50 turns.
Side AC is 90 degrees to the magnetic field. I used F=lIB
So F = 0.025 x 1.5 x 0.25 = 9.375 x 10^-3
But the answers divided it by 50 and I have no idea why you would can someone please explain? (the answer is 1.875 x 10^-4)
I also don't get part d) Predict the force on side AC of the loop if the strength of the magnetic field was halved.
The diagram shows a horizontal 10cm square loop carrying 1.5 A in a magnetic field of strength 0.25T
c) Predict the force on side AC of the loop if it was made into a coil with 50 turns.
Side AC is 90 degrees to the magnetic field. I used F=lIB
So F = 0.025 x 1.5 x 0.25 = 9.375 x 10^-3
But the answers divided it by 50 and I have no idea why you would can someone please explain? (the answer is 1.875 x 10^-4)
I also don't get part d) Predict the force on side AC of the loop if the strength of the magnetic field was halved.
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