a charged particle of mass 0.0050 kg is subjected to a 5 t magnetic field which acts at a right angle to its motion. if the particle moves in a circle of radius 0.2 m at a speed of 4 m/s , what is the magnitude of the charge on the particle?



Answer :

Magnitude of the charge on the particle is 0.02 C

What is a magnetic field?

A changing electric field in which a magnetic field, a vector field near a magnet, an electric current, or a magnetic force is observable. A magnetic field like the earth aligns a magnetic compass needle or other permanent magnet with it. Magnetic fields cause charged particles to move in circular or spiral paths. This force - applied to the current in wires in a magnetic field - underlies the operation of electric motors.

Applying,

F = qvBsinФ

Where F = Force on the charged particle

q = charge on the particle

v = velocity

B = magnetic field

Ф =  angle

Since the charged particle moves in a circle,

F = mv²/r

Where m = mass of the particle, v = velocity of the particle, r = radius of the  circle

So,

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)

Given: m = 0.005 kg

v = 4 m/s

r = 0.2 m

B = 5 T

Ф = 90° (Act at right angle)

Now, substitute the values in above equation:

q = (0.005×4)/(0.2×5×sin90°)

q = 0.02 C

To know more about magnetic field visit:

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