An ion having charge +6ee is traveling horizontally to the left at 9.00 km/skm/s when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of 5.94×10−155.94×10−15 NN. What is the magnitude of this field? Express your answer in teslas.
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Given: - \( F = 5.94 \times 10^{-15} \, \text{N} \) - \( v = 9.00 \, \text{km/s} = 9000 \, \text{m/s} \) - The charge \( q = +6e \), where \( e \) (the elementary charge) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). Show more…
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An ion having charge +6e is traveling horizontally to the left at 8.00 km/s when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of 6.94×10^(-15) N. Part B What is the magnitude of this field? Express your answer in teslas. B =
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An ion having charge $+6 e$ is traveling horizontally to the left at 8.50 $\mathrm{km} / \mathrm{s}$ when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of $6.94 \times 10^{-15} \mathrm{N} .$ What are the direction and magnitude of this field? Illustrate your method of solving this problem with a diagram.
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