Magnitude of Magnetic force on Wire

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SUMMARY

The magnitude of the magnetic force on a wire segment of length L=4.50 m carrying a current of i=35.0 A at an angle of theta=50.3 degrees in a magnetic field of magnitude B=6.70E-2 T is calculated using the formula F=ILxBsin(theta). The correct calculation yields a force of 8.12 N, confirming the user's result. The discrepancy with the book's answer, which states 11.8 N, indicates a potential error in the textbook.

PREREQUISITES
  • Understanding of the formula for magnetic force on a current-carrying wire
  • Knowledge of vector cross products in physics
  • Familiarity with trigonometric functions, specifically sine
  • Basic concepts of magnetic fields and current
NEXT STEPS
  • Review the derivation of the magnetic force formula F=ILxBsin(theta)
  • Study vector cross product calculations in physics
  • Explore the effects of angle on magnetic force in different configurations
  • Investigate common errors in physics textbooks regarding magnetic force calculations
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Physics students, educators, and anyone interested in understanding the principles of electromagnetism and magnetic forces on conductors.

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An isolated segment of wire of length L=4.50 m carries a current of magnitude i=35.0 A at an angle theta=50.3 degrees with respect to a constant magnetic field with magnitude B= 6.70E-2 T. What is the magnitude of the magnetic force on the wire?
a) 2.66 N
b) 3.86
c) 5.60 N
d) 8.12 N
e) 11.8 N
What I did was:
F=ILxBsin(theta)
=(35.0A)(4.50m)*( 6.70E-2T * sin(50.3 degrees))
I got 8.12 N but I checked on the back of the book and it says that the answer is (e) but I don't know how to get that.
 
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Your answer is OK
 
I agree that 8.12 N is correct. It looks like the book has an error here.
 

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