Solution:
For the wire to be suspended, the upward magnetic force must equal gravity: \(I L B = mg ⇒ I = \frac{mg}{LB}\). Plugging in the values: \(I = \frac{0.250 \times 9.8}{2 \times 0.7} = 1.75\text{ A}\).
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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