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A straight wire of length \( 50 \mathrm{~cm} \) carrying a current of \( 2.5 \mathrm{~A} \) is suspended in mid-air by a uniform
magnetic field of \( 0.5 \mathrm{~T} \) (as shown in figure). The mass of the wire is \( \left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right) \)
Options:
magnetic field of \( 0.5 \mathrm{~T} \) (as shown in figure). The mass of the wire is \( \left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right) \)
Solution:
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Verified Answer
The correct answer is:
\( 62.5 \mathrm{gm} \)
Given, the wire is suspended in mid air by uniform magnetic field. Now,
\[
\begin{array}{l}
\Rightarrow F=m g \\
\Rightarrow I B l=m g \\
\Rightarrow m=\frac{I B l}{g} \\
\Rightarrow m=\frac{2.5 \times 0.5 \times\left(50 \times 10^{-2}\right)}{10}=0.625 \mathrm{~kg}=62.5 \mathrm{~g}
\end{array}
\]
Thus, mass of wire is \( 62.5 \mathrm{~g} \).
\[
\begin{array}{l}
\Rightarrow F=m g \\
\Rightarrow I B l=m g \\
\Rightarrow m=\frac{I B l}{g} \\
\Rightarrow m=\frac{2.5 \times 0.5 \times\left(50 \times 10^{-2}\right)}{10}=0.625 \mathrm{~kg}=62.5 \mathrm{~g}
\end{array}
\]
Thus, mass of wire is \( 62.5 \mathrm{~g} \).
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