91Ó°ÊÓ

$\begin{array}{c}{\text{μ}}_{\text{Fe}}=2.1×{10}^{−23}\text{J}/\text{T}\\ \text{A}=1.0{\text{cm}}^2\\ \text{B}=1.5\text{T}\\ \text{ÒÏ}=7.9\text{g}/{\text{cm}}^3\end{array}$

We can find the number of atoms in iron atom by using the mass of the iron bar, the molar mass and Avogadro’s number. Then, find the dipole moment of the bar using the dipole of moment of the atom and the number of atoms in the iron bar. Torque can be calculated by using the dipole moment of the bar and the magnetic field.

Formula:

Number of atoms in the iron bar is$\mathbf{N}\mathbf{=}\mathbf{\left(}\frac{\mathbf{m}}{\mathbf{M}}\mathbf{\right)}\mathbf{×}{\mathbf{N}}_{\mathbf{A}}$

Here,

$\mathbf{m}\mathbf{}\mathbf{-}$Mass of iron bar.

$\mathbf{M}\mathbf{-}\mathbf{molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{iron}\mathbf{.}$

${\mathbf{N}}_{\mathbf{A}}\mathbf{-}\mathbf{}\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{}\mathbf{number}\mathbf{.}$

Dipole moment of the iron bar:$\mathbf{μ}\mathbf{=}{\mathbf{μ}}_{\mathbf{Fe}}\mathbf{×}\mathbf{N}$

${\mathbf{μ}}_{\mathbf{Fe}}\mathbf{-}\mathbf{}\mathbf{dipole}\mathbf{}\mathbf{moment}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{atoms}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{iron}\mathbf{}\mathbf{bar}\mathbf{.}$

(a)

$\begin{array}{c}\mathrm{Mass}\left(\mathrm{m}\right)=\mathrm{Density}\left(\mathrm{ÒÏ}\right)×\mathrm{volume}\left(\mathrm{V}\right)\\ \mathrm{m}=7.9×1.0×5.0\\ =39.5\mathrm{g}\end{array}$

The number of atoms in the iron bar is

$\begin{array}{c}\mathrm{N}=\left(\frac{\mathrm{m}}{\mathrm{M}}\right)×{\mathrm{N}}_{\mathrm{A}}\\ =\frac{39.5}{55.847}×6.023×{10}^{23}\\ =4.3×{10}^{23}\mathrm{atoms}\end{array}$

The dipole moment of the iron bar:

$\begin{array}{c}\mathrm{μ}={\mathrm{μ}}_{\mathrm{Fe}}×\mathrm{N}\\ =2.1×{10}^{-23}×(4.26×{10}^{23})\\ =8.9\mathrm{A}â‹\dots {\mathrm{m}}^2\end{array}$

The dipole moment of the bar is$\mathrm{μ}=8.9\mathrm{A}â‹\dots {\mathrm{m}}^2$

(b)

$\begin{array}{c}\mathrm{Ï„}=\text{μµþ²õ¾±²Ô}\left(\text{θ}\right)\\ =8.9×1.5×\sin \left(90\right)\\ =13\mathrm{N}.\mathrm{m}\end{array}$

The torque required is $\mathrm{T}=13\text{N.m}$