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QJEEM24Phd03d: JEE Main 2024 Physics - Class 12 JEEM24Phd03d Solution: A charge of is moving with a velocity of along the positive…

A charge of 4.0μC4.0 \mu \mathrm{C} is moving with a velocity of 4.0×106 ms−14.0 \times 10^6 \mathrm{~ms}^{-1} along the positive yy axis under a magnetic field B⃗\vec{B} of strength (2k^)T(2 \hat{k}) \mathrm{T} . The force acting on the charge is xi^Nx \hat{i} N . The value of xx is __________.

Step-by-Step Verified Solution

qamp;=4μC,v→=4×106j^m/sB→amp;=2kT^F→amp;=q(v→×B→)amp;=4×10−6(4×106j^×2k^)amp;=4×10−6×8×106i^F→amp;=32i^Nxamp;=32\begin{aligned}\mathrm{q} & =4 \mu \mathrm{C}, \overrightarrow{\mathrm{v}}=4 \times 10^6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} \\\overrightarrow{\mathrm{B}} & =2 \hat{\mathrm{k} T} \\\overrightarrow{\mathrm{F}} & =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\& =4 \times 10^{-6}\left(4 \times 10^6 \hat{\mathrm{j}} \times 2 \hat{\mathrm{k}}\right) \\& =4 \times 10^{-6} \times 8 \times 10^6 \hat{\mathrm{i}} \\\overrightarrow{\mathrm{F}} & =32 \hat{\mathrm{i}} \mathrm{N} \\\mathrm{x} & =32\end{aligned}

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