Consider a wire of length L = 0.30 m that runs north-south on a horizontal surface. There is a current of I = 0.50 A flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not shown.) The Earth's magnetic field at this location has a magnitude of 0.50 { m gauss} (or, in SI units, 0.5 imes 10^{-4} \; m tesla) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that 1\;{ m T} = 1\;{ m N / ( m A \cdot m m)}

Question

Consider a wire of length L = 0.30 m that runs north-south on a horizontal surface. There is a current of I = 0.50 A flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not shown.) The Earth's magnetic field at this location has a magnitude of 0.50 {m gauss} (or, in SI units, 0.5 	imes 10^{-4} \; m tesla) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that 1\;{m T} = 1\;{m N / (m A \cdot m m)}

Anonymous · Accepted Answer

This question is actually more simple than it seems. You just use the equation F=ILBsin(theta) with theta in radians. The answer is 4.6*10^(-6)N

helper · Answer

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