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A charged particle of mass 0.0040 kg is subjected to a 5.0-T magnetic field which acts at a right angle to its motion. If the particle moves in a circle of radius 0.20 m at a speed of 2.0 m/s, what is the magnitude of the charge on the particle?

A. 0.0080 C
B. 1300 C
C. 130 C
D. 0.00080 C

User Cicakman
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1 Answer

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Final answer:

The magnitude of the charge on the charged particle moving in a circle in a magnetic field can be found by equating magnetic force to centripetal force. Using the given values of mass, velocity, magnetic field strength, and the radius of the circle, we find the charge to be 0.0080 C.

Step-by-step explanation:

To determine the magnitude of the charge on the particle moving in a magnetic field, we can use the formula for the magnetic force that causes a charged particle to move in a circle:

F = qvB

Where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength. The force that keeps the particle in circular motion is centripetal force, which is given by the formula:

Fc = m·v²/r

Where Fc is the centripetal force, m is the mass of the particle, v is the velocity, and r is the radius of the circle. Since the magnetic force is providing the centripetal force:

qvB = m·v²/r

To find the charge q, we can rearrange the equation and solve:

q = (m·v) / (B·r)

Now plugging in the given values:

q = (0.0040 kg · 2.0 m/s) / (5.0 T · 0.20 m) = 0.0080 C

Therefore, the magnitude of the charge on the particle is 0.0080 C, which corresponds to option A.

User Sophie Alpert
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