Question

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

Answer 1

Correct Answer: 5

\[ R = \frac{\rho \ell}{A} \implies 2 \times 10^{-6} \times \frac{\ell}{10^{-5}} = 1 \implies \ell = 5 \]\[ mg = B I \ell \]\[ B = \frac{mg}{I \ell} = \frac{5}{2 \times 5} = 0.5 = 5 \times 10^{-1} \, \text{Tesla} \]

Answer 2

Step 1: Write down the given data
Current \( I = 2\,\text{A} \)
Resistance \( R = 1\,\Omega \)
Resistivity \( \rho = 2 \times 10^{-6}\,\Omega\,\text{m} \)
Area of cross-section \( A = 10\,\text{mm}^2 = 10 \times 10^{-6}\,\text{m}^2 = 1 \times 10^{-5}\,\text{m}^2 \)
Mass \( m = 500\,\text{g} = 0.5\,\text{kg} \)
Acceleration due to gravity \( g = 10\,\text{m/s}^2 \)

Step 2: Find the length of the wire
We know \( R = \rho \dfrac{L}{A} \). Hence, \[ L = \frac{R A}{\rho} = \frac{1 \times 1\times10^{-5}}{2\times10^{-6}} = 5\,\text{m}. \] Therefore, the wire length is \( L = 5\,\text{m}. \)

Step 3: Condition for magnetic suspension
For the wire to remain suspended, the magnetic force balances the weight: \[ BIL = mg. \] Hence, \[ B = \frac{mg}{IL}. \] Substitute the known values: \[ B = \frac{0.5 \times 10}{2 \times 5} = \frac{5}{10} = 0.5\,\text{T}. \] Thus, \( B = 0.5\,\text{T} = 5 \times 10^{-1}\,\text{T}. \)

Step 4: Final result
The magnitude of the magnetic field \( B \) is 5 × \(10^{-1}\,\text{T}\).

Final answer

5