Need help for DOT POINT 1.5.2 (1 Viewer)

freezeice04

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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.
 
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mathemalia

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F = nBIlsinx
since x = 90
than F1= nBIl not divided, multiplied, wierd >.<

and using the same formula, if strength was halved, then

F2 = n(B/2)Il

so the force will also be halved.
 

s2 SEductive

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T= nBIACOSx
1)
B= 0.25 T
I = 1.5 Amps
A(area) 0.1x 0.1 = 0.01m (10cm=0.1m)
n = 50 turns

T= 0.25 x 1.5 x 0.01
= 0.1875 N upwards

2)
B= 0.25/2 = 0.125 T ( divide by 2 since magnetic field is halved)
I = 1.5 Amps
A(area) 0.1x 0.1 = 0.01m (10cm=0.1m)

T= 0.125 x 1.5 x 0.01 =
= 1.875 x 10^-3 N upwards
 

airblitz7

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T= nBIACOSx
1)
B= 0.25 T
I = 1.5 Amps
A(area) 0.1x 0.1 = 0.01m (10cm=0.1m)
n = 50 turns

T= 0.25 x 1.5 x 0.01
= 0.1875 N upwards

2)
B= 0.25/2 = 0.125 T ( divide by 2 since magnetic field is halved)
I = 1.5 Amps
A(area) 0.1x 0.1 = 0.01m (10cm=0.1m)

T= 0.125 x 1.5 x 0.01 =
= 1.875 x 10^-3 N upwards
Doesn't the book have worked solutions at the back? I'm pretty sure mine does.
 

freezeice04

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T= nBIACOSx
1)
B= 0.25 T
I = 1.5 Amps
A(area) 0.1x 0.1 = 0.01m (10cm=0.1m)
n = 50 turns

T= 0.25 x 1.5 x 0.01
= 0.1875 N upwards

2)
B= 0.25/2 = 0.125 T ( divide by 2 since magnetic field is halved)
I = 1.5 Amps
A(area) 0.1x 0.1 = 0.01m (10cm=0.1m)

T= 0.125 x 1.5 x 0.01 =
= 1.875 x 10^-3 N upwards
Why are you using the Torque formula? I don't understand, it's asking for force.
 

Aquawhite

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Why are you using the Torque formula? I don't understand, it's asking for force.
Torque is the rotating force at a point on an object at a particular length. So using torque formula to calculate the required force is correct.
 

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