Magnetic Effects & Magnetism

JEE Main level, single correct, medium

133 of 276

Question 133 of 276

A velocity selector consists of electric field E⃗=Ek^\vec{E}=E\hat{k} and magnetic field B⃗=Bj^\vec{B}=B\hat{j} with B=12mTB=12mT. The value EE required for an electron of energy 728eV728eV moving along the positive xx-axis to pass undeflected is
(Given, mass of electron =9.1×10−31kg=9.1\times {10}^{-31}kg)

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Answer: (A) 192kVm−1192kV{m}^{-1}

Fiven that E⃗=Ek^\vec{E}=E\hat{k} and B⃗=12j^mT\vec{B}=12\hat{j}mT
Kinetic energy=728eV=728eV
Kinetic energy =12mv2=\dfrac{1}{2}m{v}^{2}
⇒728eV=12×9.1×10−31×v2\Rightarrow 728eV=\dfrac{1}{2}\times 9.1\times {10}^{-31}\times {v}^{2}
⇒728×1.6×10−19=12×9.1×10−31×v2\Rightarrow 728\times 1.6\times {10}^{-19}=\dfrac{1}{2}\times 9.1\times {10}^{-31}\times {v}^{2}
⇒v=16×106ms−1\Rightarrow v=16\times {10}^{6}m{s}^{-1}
For electron to move undeflected net force on it should be zero.
⇒eE=evB\Rightarrow eE=evB
⇒E=vB=16×106×12×10−3\Rightarrow E=vB=16\times {10}^{6}\times 12\times {10}^{-3}
⇒E=192×103Vm−1=192kVm−1\Rightarrow E=192\times {10}^{3}V{m}^{-1}=192kV{m}^{-1}

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