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, \( g = 10 \, \text{m/s}^2 \)).

Updated On: Nov 21, 2024
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Correct Answer: 5

Solution and Explanation

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

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