A charged particle enters a uniform magnetic field with velocity v0=4 m/s perpendicular to it, the length of magnetic field is x=32R, where R is the radius of the circular path of the particle in the field. Find the magnitude of change in velocity (in m/s ) of the particle when it comes out of the field.

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answer is 4.

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Detailed Solution

The particle will come out of the magnetic field at an angle θ=600with the original direction.Δv→=v0cos60i^+v0sin60j^-v0i^=-v022+v0322 ⇒|Δv→|=v0

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