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2 years ago

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A particle of mass 10 g and charge 72 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is

−9.8j m/s2. The velocity of the particle is a constant 19i hat km/s, which is perpendicular to the magnetic field. What, then, is the magnetic field? (Express your answer in vector form.)

1 answer:

sineoko [7]2 years ago

5 0

Answer:

$-0.07163\hat k\ T$ or 0.07163 T into the page

Explanation:

m = Mass of particle = 10 g

a = Acceleration due to gravity = -9.8j m/s²

v = Velocity of particle = 19i km/s

q = Charge of particle = 72 μC

B = Magnetic field

Here the magnetic and gravitational forces on the particle are applied in the opposite direction so,

$F_b=F_g$

$F_b=qvBsin\theta\\\Rightarrow F_b=qvBsin90\\\Rightarrow F_b=72\times 10^{-6}\times 19000B$

$F_g=ma\\\Rightarrow F_g=0.01\times -9.8$

$72\times 10^{-6}\times 19000B=0.01\times -9.8\\\Rightarrow B=\frac{0.01\times -9.8}{72\times 10^{-6}\times 19000}\\\Rightarrow B=-0.07163\hat k\ T$

The magnetic field is 0.07163 T into the page

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Explanation:

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The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

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