Μ nia τ μ b μ b sin θ μ ni π r 2 μ 30 5a π

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μ = NIA τ = μ × B = μ B sin θ μ = NI π r 2 ( ) μ = 30 ( ) 5A ( ) π 0.05m ( ) 2 = 1.18 A m 2 τ = 1.18 A m 2 ( ) 1.20T ( ) sin90 ° τ = 1.42 N m For torque:
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Torque A Current Loop Answer For direction, look at four separate points on the circular loop (top, left, bottom, and right parts): Using RHR on the top and bottom gives us: Zero magnetic force (I and B are parallel in both instances). B field Using RHR on the left gives us: A magnetic force that is out of board. Using RHR on the right gives us: A magnetic force that is into the board. So the coil will flip such that the right side goes into the board and the left side comes out of the board.
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Magnetic Fields Back to our two step model of calculating the magnetic force. Chapter 29 concentrates on calculating the magnetic field due to moving charges (current). In general, for a current carrying wire this becomes a very difficult calculation. We will start with the easiest case (a long straight wire) and move on to tougher cases.
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Magnetic Fields The magnitude of the magnetic field at a perpendicular distance r from a long, straight wire carrying a current of I is: where μ o is called the permeability of free space and has the value:
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Magnetic Fields The direction of the magnetic field from the wire will be given by another version of the Right Hand Rule. Grasp the wire in your right hand. Point your thumb in the direction of the current I (positive charge flow). Your fingers will curl in the direction of the magnetic field.
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Step 1 Many times you will be asked for the magnetic field at a given point.
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