Question

What magnetic field strength and direction will levitate the $2.0\mathrm{g}$wire in FIGURE EX29.33?

Short answer

The magnetic field in the wire is $0.13\mathrm{T}$and it is out of the paper.

Step-by-step solution

Given information

We have given,

Current = $1.5\mathrm{A}$

Magnetic field region =$10\mathrm{cm}$

We have to find the strength of magnetic field.

Simplify

We know the magnetic force is

$\overrightarrow{\mathrm{F}}=i(\overrightarrow{L}\times \overrightarrow{B})$

To simplify it let us consider the magnetic field is out of the paper

$$\begin{gathered} \mathrm{F}=\mathrm{iLB}\sin {\left(90\right)}^0=\mathrm{iLB} \\ \mathrm{F}=\mathrm{ILB} \end{gathered}$$

This force is acting upward and it is balancing the weight of the wire. then,

$$\begin{gathered} \mathrm{iLB}=\mathrm{mg} \\ (1.5\mathrm{A})\times (0.1\mathrm{m})\times \mathrm{B}=(0.002\mathrm{Kg})\times 10 \\ \mathrm{B}=0.13\mathrm{T} \end{gathered}$$

and the direction of magnetic field is out of the paper then only the force due to magnetic field will be able to balance the weight.