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Debora [2.8K]
3 years ago
12

What is the force exerted on a charge of 2.5 µC moving perpendicular through a magnetic field of 3.0 × 102 T with a velocity of

5.0 × 103 m/s?
3.8 N
38 N
3.8 × 105 N
3.8 × 106 N
Physics
2 answers:
OverLord2011 [107]3 years ago
8 0

Given:

B = 3 \times 10^{2} T

V=5 \times 10^{3} \frac{m}{s}

q = 2.5 × 10^{-6} C

α = 90

To find:

Force = ?

Formula used:

Force on the moving charge is given by,

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

Solution:

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

F = 2.5 \times 10^{-6} \times 3 \times 10^{2}  \times 5 \times 10^{3}

F = 37.5 × 10^{-1}

F = 3.75 Newton

Thus, the force acting on the moving charge is 3.75 Newton.

tangare [24]3 years ago
6 0
It's 3.8 N - just took it online and got it right. 
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