Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

117 of 276

Question 117 of 276

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms−12 \times 10^5 \mathrm{~ms}^{-1}. When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is x×104 N/Cx \times 10^4 \mathrm{~N} / \mathrm{C}. The value of xx is _______ Take the mass of the proton =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg}.

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Answer: 2

Bvq=EqE=Bvr=vmBqB=mvrqE=(mvrq)v=mv2rq=1.6×10−27×4×10102×10−2×1.6×10−19=2×104 N/Cx=2\begin{aligned} & B v q=E q \\ & E=B v \\ & r=\frac{v m}{B q} \\ & B=\frac{m v}{r q} \\ & E=\left(\frac{m v}{r q}\right) v=\frac{m v^2}{r q} \\ & =\frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{2 \times 10^{-2} \times 1.6 \times 10^{-19}} \\ & =2 \times 10^4 \mathrm{~N} / \mathrm{C} \\ & x=2\end{aligned}

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