A wire that is vertical and straight carries a current 1.20 A downward in a region b/t the poles of a large superconducting electromagnet,where the magnetic filed has magnitude B = 0.588 T and is horizontal. What are the size and direction of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is a) east? b)south? c) 30 degrees south of west?According to the solutions, the force remains constant for parts&nbsp;a,b,c:F=I⋅l⋅B=7.06×10−3&nbsp;but for part c,why isn't the 30 degree angle used to calculate the force?

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A wire that is vertical and straight carries a current 1.20 A downward in a region b/t the poles of a large superconducting electromagnet,where the magnetic filed has magnitude B = 0.588 T and is horizontal. What are the size and direction of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is a) east? b)south? c) 30 degrees south of west?According to the solutions, the force remains constant for parts&amp;nbsp;a,b,c:F=I⋅l⋅B=7.06×10−3&amp;nbsp;but for part c,why isn&#039;t the 30 degree angle used to calculate the force?

yagombyeR · Accepted Answer

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Jeffrey Jordon · Accepted Answer

The equation for force on the wire is,F=BILsin⁡θWhere&amp;nbsp;θ&amp;nbsp;is the angle between magnetic field and current.The current direction is downward as mentioned in the question(a)The direction of a magnetic field is eastwardSo, the angle between a magnetic field and current isθ=90∘Therefore,F=0.588×1.2×10−2×sin⁡(90)=7.056×10−3NAccording to Flemings left-hand rule, the direction of force is into or out of the pageb)The direction of a magnetic field is in the south.The angle between a magnetic field and current is the angle between the direction of the magnetic field and the direction of the current.θ=0∘Therefore,F=0.588×1.2×10−2×sin⁡(0)=0c)The direction of magnetic field is&amp;nbsp;30∘&amp;nbsp;south of west.So, the angle between magnetic field and current isθ=30∘Therefore,F=0.588×1.2×10−2×sin⁡(30)=3.528×10−3NThe direction of force is,&amp;nbsp;90−30=60∘&amp;nbsp;south of west into the page.

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A straight vertical wire carries a current of 1.20 A downward in a region b/t the poles of a large superconducting electromagnet,where the magnetic fi

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