A long straight wire of length $2 ext{ m}$ and mass $250 ext{ g}$ is suspended horizontally in a uniform h

Question

A long straight wire of length $2 	ext{ m}$ and mass $250 	ext{ g}$ is suspended horizontally in a uniform horizontal magnetic field of $0.7 	ext{ T}$. The amount of current flowing through the wire will be: ($g=9.8 	ext{ m s}^{-2}$)

Accepted Answer

1.  **Identify Condition for Equilibrium:** For the wire to be suspended (levitated) horizontally, the upward magnetic force ($F_m$) must balance the downward gravitational force (weight, $W$).
    $$F_m = W$$
2.  **Formulate Equations:**
    *   Weight: $W = mg$
    *   Magnetic Force: $F_m = B I L \sin	heta$. For the most efficient suspension, the field is perpendicular to the wire, so $	heta = 90^{\circ}$ and $\sin(90^{\circ}) = 1$. Thus, $F_m = B I L$.
    *   Equilibrium Equation: $B I L = mg$
3.  **Rearrange for Current ($I$):**
    $$I = \frac{mg}{B L}$$
4.  **Convert Units and Substitute:**
    *   Mass ($m$) = $250 	ext{ g} = 0.25 	ext{ kg}$
    *   Length ($L$) = $2 	ext{ m}$ (inferred from context and calculation match)
    *   Magnetic Field ($B$) = $0.7 	ext{ T}$
    *   Gravity ($g$) = $9.8 	ext{ m/s}^2$
    $$I = \frac{0.25 	imes 9.8}{0.7 	imes 2}$$
5.  **Calculate:**
    $$I = \frac{2.45}{1.4} = 1.75 	ext{ A}$$

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