A 2 A current carrying straight metal wire of resistance 1 , resistivity 2 × 10^ -6 m , area of cross-section 10 ~mm^2 and mass 500 ~g is suspended horizontally in mid air by applying a uniform magnetic field B . The

Correct answer: 0 aligned & R= A 2 × 10^ -6 × 10^ -5 =1 =5 & mg=Bi aligned B= mg i =(5)/(2 × 5)=0.5=5 × 10^ -1 Tesla

JEE Main 20245 AprilMorning Shift

JEE Main 2024 · 5 April · Morning Shift · Physics · Magnetic Effects of Current

A 2 A current carrying straight metal wire of resistance 1 ,...

A 2 A current carrying straight metal wire of resistance

1 \Omega

, resistivity

2 \times 10^{-6} \Omega \mathrm{m}

, area of cross-section

10 \mathrm{~mm}^2

and mass

500 \mathrm{~g}

is suspended horizontally in mid air by applying a uniform magnetic field

\vec{B}

. The magnitude of

B

is _____

\times 10^{-1} \mathrm{~T}

(given,

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2

\begin{aligned} & \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \Rightarrow \frac{2 \times 10^{-6} \times \ell}{10^{-5}}=1 \Rightarrow \ell=5 \\ & \mathrm{mg}=\mathrm{Bi} \ell\end{aligned}

\mathrm{B}=\frac{\mathrm{mg}}{\mathrm{i} \ell}=\frac{5}{2 \times 5}=0.5=5 \times 10^{-1} \mathrm{Tesla}$