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Question 17 of 17 View Policies Current Attempt in Progress The coil in the figure carries current $i = 2.31$ A in the direction indicated, is parallel to an xz plane, has 3 turns and an area of $5.46 \times 10^{-3}$ $m^2$, and lies in a uniform magnetic field $\vec{B} = (1.62\hat{i} + 2.35\hat{j} - 4.63\hat{k})$ mT. (a) What is the magnetic potential energy of the coil-magnetic field system? (b) What is the magnetic torque (in unit-vector notation) on the coil?

          Question 17 of 17
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Current Attempt in Progress
The coil in the figure carries current $i = 2.31$ A in the direction indicated, is parallel to an xz plane, has 3 turns and an area of $5.46 \times 10^{-3}$ $m^2$, and lies in a uniform magnetic field $\vec{B} = (1.62\hat{i} + 2.35\hat{j} - 4.63\hat{k})$ mT. (a) What is the magnetic potential energy of the coil-magnetic field system? (b) What is the magnetic torque (in unit-vector notation) on the coil?
        
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Question 17 of 17
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Current Attempt in Progress
The coil in the figure carries current i = 2.31 A in the direction indicated, is parallel to an xz plane, has 3 turns and an area of 5.46 × 10^-3 m^2, and lies in a uniform magnetic field B⃗ = (1.62î + 2.35ĵ - 4.63k̂) mT. (a) What is the magnetic potential energy of the coil-magnetic field system? (b) What is the magnetic torque (in unit-vector notation) on the coil?

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Question 17 of 17 The coil in the figure carries current i = 2.31 A in the direction indicated, is parallel to the xz plane, has 3 turns and an area of 5.46 * 10^3 m^2, and lies in a uniform magnetic field B = 1.62 * 10^3 - 2.35 * 10^3 - 4.63 * 10^-3 T. (a) What is the magnetic potential energy of the coil-magnetic field system? (b) What is the magnetic torque (in unit-vector notation) on the coil?
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Transcript

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00:01 Hello students in the question we are given current i it is equal to 2 .10 ampere number of turns equal to 6 and area across section equal to 3 .86 into 10 to the power minus 3 meter square here we have magnetic field b it is equal to 1 .56 i cap minus 3 .21 j cap minus 4 .35 k cap in units of micro tesla here we have formula magnetic moment m it is equal to number of turns into current i into area of cross section so here we can write this is equal to m magnetic moment it is equal to number of turns 6 into current 2 .10 into area 3 .86 into 10 to the power minus 3 now axis of this loop it is in positive y direction so we can write j cap in positive y direction so this comes out to be equal to 48 .636 into 10 to the power minus 3 ampere meter square in direction j cap for part a we have magnetic potential energy of coil it is given by formula magnetic potential energy u it is equal to minus m magnetic moment dot product of magnetic field so here we can write this is equal to minus 48 .636 into 10 to the power minus 3 j cap into 1 .56 i cap minus 3 .21 j cap minus 4 .35 k cap in units of micro tesla in here…
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