Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

2.45 A

2.25 A

2.75 A

1.75 A

Step-by-Step Solution

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, $W$ ). $F_m = W$
Formulate Equations: Weight: $W = mg$ Magnetic Force: $F_m = B I L \sin\theta$ . For the most efficient suspension, the field is perpendicular to the wire, so $\theta = 90^{\circ}$ and $\sin(90^{\circ}) = 1$ . Thus, $F_m = B I L$ .

Equilibrium Equation: $B I L = mg$

Rearrange for Current ( $I$ ): $I = \frac{mg}{B L}$
Convert Units and Substitute: Mass ( $m$ ) = $250 \text{ g} = 0.25 \text{ kg}$ Length ( $L$ ) = $2 \text{ m}$ (inferred from context and calculation match) Magnetic Field ( $B$ ) = $0.7 \text{ T}$ Gravity ( $g$ ) = $9.8 \text{ m/s}^2$ $I = \frac{0.25 \times 9.8}{0.7 \times 2}$
Calculate: $I = \frac{2.45}{1.4} = 1.75 \text{ A}$

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