Question

A straight horizontal copper wire has a current $I=28\text{A}$ flowing through it. What is the mimimum magnitude of magnetic field required to suspend the wire in air by balancing the gravitational force acting on it?
Given: the linear density of wire is $46.6\text{g/m}$

• A. $1.6\times {10}^{-2}\text{T}$
• B. $2.48\times {10}^{-3}\text{T}$
• C. $3.2\times {10}^{-2}\text{T}$
• D. $4.2\times {10}^{-2}\text{T}$

Answer 1

The correct option is A $1.6\times {10}^{-2}\text{T}$

The wire is horizontal, hence the $\overrightarrow{L}$ is directed horizontally i.e let it be along the $+vey$-axis in the $xy$ (horizontal plane).

To balance the weight of wire (acting vertically downward), the direction of magnetic force must be vertically upwards.

Thus using, $\overrightarrow{F}=I(\overrightarrow{L}\times \overrightarrow{B})$, we can say that direction of magnetic field must be along +ve x-axis.

Let the angle between $\overrightarrow{L}$ & $\overrightarrow{B}$ be $\theta$, thus magnitude of magnetic force on wire,

$F=BIl\sin \theta$

Using equilibrium condition;

$BIL\sin \theta =mg$

or, $B=\frac{mg}{IL\sin \theta }$

For $B$ to be minimum, $\sin \theta$ should be maximum i.e $\sin \theta =1$ or $\theta ={90}^o$




$\Rightarrow B_{min}=\frac{mg}{IL}$

Here,

$\frac{m}{L}=46.6g/m=46.6\times {10}^{-3}\text{kg/m}$

$B_{min}=\frac{46.6\times {10}^{-3}\times 9.8}{28}$

$\therefore B_{min}=1.6\times {10}^{-2}\text{T}$