Verified Solution • 2025–26 Session moderate

QJEEM24Ph5b2e: JEE Main 2024 Physics - Class 12 JEEM24Ph5b2e Solution: An electron moves through a uniform magnetic field . At a…

An electron moves through a uniform magnetic field B⃗=B0i^+2B0j^T\vec{B}=B_0 \hat{i}+2 B_0 \hat{j} T . At a particular instant of time, the velocity of electron is u⃗=3i^+5j^ m/s\vec{u}=3 \hat{i}+5 \hat{j} \mathrm{~m} / \mathrm{s} . If the magnetic force acting on electron is F⃗=5ek^N\vec{F}=5 e \hat{k} N , where ee is the charge of electron, then the value of B0B_0 is _________ TT .

Step-by-Step Verified Solution

amp;F→=q(v→×B→)amp;5ek^=e(3i^+5j^)×(B0i^+2 B0j^)amp;5ek^=e(6 B0k^−5 B0k^)amp;⇒B0=5 T\begin{aligned}& \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\& 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \times\left(\mathrm{B}_0 \hat{\mathrm{i}}+2 \mathrm{~B}_0 \hat{\mathrm{j}}\right) \\& 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}\left(6 \mathrm{~B}_0 \hat{\mathrm{k}}-5 \mathrm{~B}_0 \hat{\mathrm{k}}\right) \\& \Rightarrow \mathrm{B}_0=5 \mathrm{~T}\end{aligned}

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