Problem

What is the force on a $1.38 C$ charge moving perpendicular to a $6.12 T$ magnetic field at $7.82 m / s$ ?
$1.76 N$
$34.7 N$
$0.0289 N$
$66.0 N$

Answer

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Answer

$F = 66.014 N$

Steps

Step 1 ： $F = q v B \sin θ$

Step 2 ： $F = (1.38 C) (7.82 \frac{m}{s}) (6.12 T) \sin (\frac{π}{2})$

Step 3 ： $F = 66.014 N$

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