A 2A current carrying straight metal wire of resistance 1 Ω, resistivity 2 × 10^–6 Ωm, area of cross-section 10 mm² and mass 500 g is suspended horizontally in mid air by applying a uniform magnetic field B. The magnitude of B is ............. × 10^–1 T (given, g = 10 m/s²)

The correct option/answer is 5 (which means the value to be filled in the blank before $\times {10}^{-1}\text{T}$ is 5, giving a magnetic field magnitude $B=5\times {10}^{-1}\text{T}=0.5\text{T}$).

Step-by-Step Derivation and Explanation:

1. Identify the given parameters:
Current in the wire, $I=2\text{A}$
Resistance of the wire, $R=1\Omega$
Resistivity of the metal, $\rho =2\times {10}^{-6}\Omega \text{m}$
Area of cross-section, $A=10{\text{mm}}^2=10\times {10}^{-6}{\text{m}}^2={10}^{-5}{\text{m}}^2$
Mass of the wire, $m=500\text{g}=0.5\text{kg}$
Acceleration due to gravity, $g=10{\text{m/s}}^2$

2. Determine the length of the wire:
The relation between resistance $R$, resistivity $\rho$, length $l$, and cross-sectional area $A$ of a wire is given by:

$R=\frac{\rho l}{A}$

Rearranging the formula to solve for the length $l$:

$l=\frac{R\cdot A}{\rho }$

Substitute the values into the equation:

$l=\frac{1\times 10\times {10}^{-6}}{2\times {10}^{-6}}=5\text{m}$

3. Condition for suspension in mid-air:
For the wire to be suspended horizontally in mid-air, the upward magnetic force acting on the wire must balance its downward gravitational force (weight).

$F_m=F_g$

The gravitational force is:

$F_g=m\cdot g$

The maximum magnetic force on a straight current-carrying wire of length $l$ in a uniform magnetic field $B$ (with the field perpendicular to the wire) is:

$F_m=B\cdot I\cdot l$

Equating the two forces:

$B\cdot I\cdot l=m\cdot g$

4. Calculate the magnetic field magnitude B:
Solve for $B$:

$B=\frac{m\cdot g}{I\cdot l}$

Substitute the values:

$B=\frac{0.5\times 10}{2\times 5}=\frac{5}{10}=0.5\text{T}$

Expressing the magnetic field in the form $B=x\times {10}^{-1}\text{T}$:

$0.5\text{T}=5\times {10}^{-1}\text{T}$

Thus, the magnitude of the magnetic field multiplier is 5.