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https://texercises.com/exercise/magnetic-force-on-current/
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Exercise:
A wire with length LO running from left to right direction of the positive x axis carries a current of IO. It is placed in a homogeneous magnetic field given by the components B_xBxO B_yByO B_zBzO. abcliste abc Calculate the force acting on the wire using the vector version for the magnetic force. abc Determine the angle between the direction of the current and the direction of the magnetic field and calculate the force using the scalar version for the magnetic force. abcliste

Solution:
abcliste abc The force vector is given by vecF I vecLcrossvecB I leftmatrix Lmatrixrightcross leftmatrix B_xB_yB_zmatrixright Ileftmatrix --L B_zL B_y-matrixright I Lleftmatrix - B_z B_ymatrixright The magnitude of the force is therefore F absvecB FF labelsolutiona ItimesLtimessqrtBz^+By^ F approx resultFP- abc The cosine of the angle between vecL and vecB can be found using the dot product: costheta fracvecLvecBabsvecLabsvecB fracL B_xLsqrtB_x^+B_y^+B_z^ fracB_xsqrtB_x^+B_y^+B_z^ Since cos^theta + sin^theta the sine of the angle is sheta sqrt-cos^theta sqrt-fracB_x^B_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ Using the scalar version for the magnetic force we thus find F I L B sheta I L sqrtB_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ I L sqrtB_y^+B_z^ This is the same expression as refsolutiona so both versions lead to the same result. abcliste
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Exercise:
A wire with length LO running from left to right direction of the positive x axis carries a current of IO. It is placed in a homogeneous magnetic field given by the components B_xBxO B_yByO B_zBzO. abcliste abc Calculate the force acting on the wire using the vector version for the magnetic force. abc Determine the angle between the direction of the current and the direction of the magnetic field and calculate the force using the scalar version for the magnetic force. abcliste

Solution:
abcliste abc The force vector is given by vecF I vecLcrossvecB I leftmatrix Lmatrixrightcross leftmatrix B_xB_yB_zmatrixright Ileftmatrix --L B_zL B_y-matrixright I Lleftmatrix - B_z B_ymatrixright The magnitude of the force is therefore F absvecB FF labelsolutiona ItimesLtimessqrtBz^+By^ F approx resultFP- abc The cosine of the angle between vecL and vecB can be found using the dot product: costheta fracvecLvecBabsvecLabsvecB fracL B_xLsqrtB_x^+B_y^+B_z^ fracB_xsqrtB_x^+B_y^+B_z^ Since cos^theta + sin^theta the sine of the angle is sheta sqrt-cos^theta sqrt-fracB_x^B_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ Using the scalar version for the magnetic force we thus find F I L B sheta I L sqrtB_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ I L sqrtB_y^+B_z^ This is the same expression as refsolutiona so both versions lead to the same result. abcliste
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Magnetism
Tags
biot-savart, cross product, magnetic forces
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Difficulty
(3, default)
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0 (default)
Language
ENG (English)
Type
Calculative / Quantity
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