Q130 / 200 · NEET 2023 (Manipur) · Physics

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

  • A
    $2.75 \mathrm{~A}$
  • Correct answer
    B
    $1.75 \mathrm{~A}$
  • C
    $2.45 \mathrm{~A}$
  • D
    $2.25 \mathrm{~A}$

Solution

Figure
we know $\begin{aligned} & F=m g \\ & I \ell B=m g\end{aligned}$ $I=\frac{m g}{\ell B}=\frac{250 \times 10^{-3} \times 9.8}{(2)(0.7)}$ $\begin{aligned} & I=1750 \times 10^{-3} \mathrm{~A} \\ & I=1.75 \mathrm{~A}\end{aligned}$

Hence, the correct answer is option: 'B'.

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