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# A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V m^{–1} makes the path straight. Find the charge/mass ratio of the particle.

Answer :

Given-

Diameter of the circle = 1.0 cm

Thus, radius of circle, *r* = = 0.5 × 10^{−2} m

Magnetic field, *B* = 0.40 T

Electric field, *E* = 200 V m^{−1}

From the question we can confer that, the particle is moving in a circle under the action of a magnetic field.

But when an electric field is applied on the particle, it moves in a straight line.

So, we can say that the electric field is balanced by the magnetic field ie,

Magnetic force, we know, Lorentz force F is given by -

where,

q = charge on an electron

v = velocity of the electron

B=magnetic field

θ= angle between B and v

The radius of the circular path described by a particle in a magnetic field r,

where,

*m* is the mass of a proton

v= velocity of the particle

B = magnetic force

q= charge on the particle = C

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