What Is the Magnitude of B for a Current-Carrying Wire in a Magnetic Field?

Thread starter Ry122
Start date May 21, 2009
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Current Force Wire

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The force on a current-carrying wire in a magnetic field is calculated using the formula F = BILsin(θ). In this scenario, a straight wire of 90 cm carrying a 60 A current at a 45-degree angle to the magnetic field experiences a force of 2.0 N. To find the magnitude of the magnetic field B, the equation is rearranged to B = F / (ILsin(θ)). Substituting the values gives B = 2.0 N / (60 A * 0.90 m * sin(45°)), resulting in a calculated magnetic field strength. The correct approach accounts for the angle between the wire and the magnetic field.

May 21, 2009

Ry122

A straight wire of length 90 cm carries a current of 60 A and makes an angle of 45 degrees with a uniform magnetic field. If the force on the wire is 2.0 N what is the magnitude of B.

my attempt:
F=BIL
2=B(6)(.90)
B=.37
What am i doing wrong?

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May 21, 2009

rock.freak667

Homework Helper

F=BIL is assuming the conductor is at 90 degrees to the magnetic field.

if it is at an angle \theta, then the formula becomes F=BILsin\theta

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