The magnetic force acting on the conductor
F=I⋅B⋅l⋅sinαF=I\cdot B\cdot l\cdot\sin \alphaF=I⋅B⋅l⋅sinα
Assume that l=1ml=1 ml=1m and α=30°\alpha=30°α=30°. So
F=I⋅B⋅l⋅sinα=3.5⋅0.1⋅1⋅sin30°=0.175NF=I\cdot B\cdot l\cdot\sin \alpha=3.5\cdot 0.1\cdot 1\cdot \sin30°=0.175 NF=I⋅B⋅l⋅sinα=3.5⋅0.1⋅1⋅sin30°=0.175N
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