A 0.40 C particle moves with a speed of 19 m/s through a region where the magnet

Question

A 0.40 C particle moves with a speed of 19 m/s through a region where the magnetic field has a strength of 0.99 T . At what angle to the field is the particle moving if the force exerted on it is 4.8×106N? Express your answer using two significant figures. At what angle to the field is the particle moving if the force exerted on it is 3.0×106N? Express your answer using two significant figures. At what angle to the field is the particle moving if the force exerted on it is 1.0×107N? Express your answer using two significant figures.

Explanation / Answer

a)

here

sin(theta) = F / (q * v * B )

theta = sin^-1((4.8 * 10^-6) / (0.4 * 10^-6 * 19 * 0.99))

theta = 39.64 deg

b)

by using the same formula

sin(theta) = F / (q * v * B )

theta = sin^-1((3 * 10^-6) / (0.4 * 10^-6 * 19 * 0.99))

theta = 23.5 deg

c)

sin(theta) = F / (q * v * B )

theta = sin^-1((1 * 10^-7) / (0.4 * 10^-6 * 19 * 0.99))

theta = 0.7615 deg

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