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QJEEM23Ph28ba: JEE Main 2023 Physics - Class 12 JEEM23Ph28ba Solution: A charge particle of accelerated by a potential difference of…

A charge particle of 2 μC2 ~\mu \mathrm{C} accelerated by a potential difference of 100 V100 \mathrm{~V} enters a region of uniform magnetic field of magnitude 4 mT4 ~\mathrm{mT} at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm3 \mathrm{~cm} inside magnetic field. The mass of the charge particle is __________ ×10−18 kg\times 10^{-18} \mathrm{~kg}

Step-by-Step Verified Solution

r=mvqB=2kmqB,r=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B},

m=r2q2B22km=\frac{r^2 q^2 B^2}{2 k}

JEE Main 2023 (Online) 1st February Morning Shift Physics - Magnetic Effect of Current Question 71 English Explanation

m=amp;1100×3100×2×2×4×10−3×4×10−3×10−122×(100)×10−6amp;=144×10−18 kg\begin{aligned}\mathrm{m}= & \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}} \\\\& =144 \times 10^{-18} \mathrm{~kg}\end{aligned}

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