A 2 A current carrying straight metal wire of resistance 1 Ω , resistivity 2 × 10 -6 Ω m , area of cross-section 10 mm 2 and mass 500 g is suspended horizontally in mid air by applying a uniform magnetic field overrightarrow B . The magnitude of B is × 10 -1 T (given, g =10 m / s 2 )

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 5 April, Shift 1 — Question 50

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 g is suspended horizontally in mid air by applying a uniform magnetic field $\overrightarrow{\mathrm{B}}$ . The magnitude of B is $\qquad$ $\times 10^{-1} \mathrm{~T}$ (given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \Rightarrow \frac{2 \times 10^{-6} \times \ell}{10^{-5}}=1 \Rightarrow \ell=5$

$\mathrm{mg}=\mathrm{Bi} \ell$

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

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Force and Torque on Wires and Loops, Magnetic Dipole Moment

In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field…

Answer: 5

Step-by-step solution

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