Magnetic Effects & Magnetism

JEE Main level, single correct, medium

164 of 276

Question 164 of 276

A particle having the same charge as of electron moves in a circular path of radius 0.5cm0.5cm under the influence of a magnetic field of 0.5T.0.5T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron =1.6×10−19C=1.6\times {10}^{-19}C )

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Answer: (C) 2.0×10−24kg2.0\times {10}^{-24}kg

Force to due to magnetic field

F=evB=mv2RF=evB=\dfrac{m{v}^{2}}{R}

with electric field also, net force is zero.

∴eE=evB∴eE=evB

⇒E=vB\Rightarrow E=vB

E=eBRm.BE=\dfrac{eBR}{m}.B

⇒m=eB2RE=1.6×10−19×(0.5)2.(0.5×10−2)100kg\Rightarrow m=\dfrac{e{B}^{2}R}{E}=\dfrac{1.6\times {10}^{-19}\times {\left(0.5\right)}^{2}.\left(0.5\times {10}^{-2}\right)}{100}kg

=2.0×10−24kg=2.0\times {10}^{-24}kg

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