Solution:
For horizontal equilibrium, the upward magnetic force balances the weight: \(I L B = m g ⇒ I = \frac{mg}{LB} = \frac{0.25 \times 9.8}{2 \times 0.7} = 1.75 \text{A}\).
A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be (\(g = 9.8 \text{ms}^{-2}\))
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