Question

A 2A current-carrying straight metal wire of resistance 1 Ω, resistivity $2 \times 10^{-6} \, \Omega \, \text{m}$, area of cross-section $10 \, \text{mm}^2$, and mass 500 g is suspended horizontally in mid-air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is ________ $\times 10^{-1} \, \text{T}$ (given, $g = 10 \, \text{m/s}^2$).

Answer 1

R = \frac{\rho \ell}{A} \implies 2 \times 10^{-6} \times \frac{\ell}{10^{-5}} = 1 \implies \ell = 5 $$$$ mg = B I \ell $$$$ B = \frac{mg}{I \ell} = \frac{5}{2 \times 5} = 0.5 = 5 \times 10^{-1} \, \text{Tesla}