Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

192 of 276

Question 192 of 276

A 2 A current carrying straight metal wire of resistance 1Ω1 \Omega, resistivity 2×10−6Ωm2 \times 10^{-6} \Omega \mathrm{m}, area of cross-section 10 mm210 \mathrm{~mm}^2 and mass 500 g500 \mathrm{~g} is suspended horizontally in mid air by applying a uniform magnetic field B⃗\vec{B}. The magnitude of BB is _____×10−1 T\times 10^{-1} \mathrm{~T} (given, g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 ).

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Answer: 5

R=ρℓA⇒2×10−6×ℓ10−5=1⇒ℓ=5mg=Biℓ\begin{aligned} & \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \Rightarrow \frac{2 \times 10^{-6} \times \ell}{10^{-5}}=1 \Rightarrow \ell=5 \\ & \mathrm{mg}=\mathrm{Bi} \ell\end{aligned}
B=mgiℓ=52×5=0.5=5×10−1Tesla\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} \ell}=\frac{5}{2 \times 5}=0.5=5 \times 10^{-1} \mathrm{Tesla}

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