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Question

Physics Question on Moving charges and magnetism

A particle of charge ee and mass mm moves with a velocity vv in a magnetic field BB applied perpendicular to the motion of the particle. The radius rr of its path in the field is

A

mvBe\frac {mv}{Be}

B

Bemv\frac {Be}{mv}

C

evBm\frac {ev}{Bm}

D

Bvem\frac {Bv}{em}

Answer

mvBe\frac {mv}{Be}

Explanation

Solution

Radius of path, r=mvqBr=\frac{m v}{q B}
Here, q=eq=e
So, r=mveBr=\frac{m v}{e B}