Chapter 32: Q50P (page 970)

A magnetic rod with length $6.00 cm$ , radius $3.00 mm$ , and (uniform) magnetization $2.7 \times 10^{3}$ A/mcan turn about its center like a compass needle. It is placed in a uniform magnetic field $\vec{B}$ of magnitude $35.0 mT$ , such that the directions of its dipole moment and make an angle of $68.0 °$ . (a) What is the magnitude of the torque on the rod due to $\vec{B}$ ? (b) What is the change in the orientation energy of the rod if the angle changes to $34.0 °$ ?

Short Answer

Expert verified

The magnitude of the torque on the rod due to $\bar{B}$ is $τ = 1.49 \times 10^{- 4} N . m$
The change in the orientation energy of the rod is $Δ U = - 72.9 \times 10^{- 6} J$

Step by step solution

Listing the given quantities

$L = 0.06 m$

$R = 3.00 \times 10^{- 3} m$

$B = 0.035 T$

The magnetization $M = 2.70 \times 10^{3} A / m$

Understanding the concepts of magnetic dipole moment

Here, we need to use the equation of torque due to magnetic field related with magnetic dipole moment and magnetic field and the equation of change in the orientation energy.

Formulae:

Torque due to magnetic field $τ = μBsin (θ)$

The change in the energy orientation: $ΔU = - μB \times (\cos (θ_{f}) - \cos (θ_{i}))$

(a) Calculations of the magnitude of torque on the rod

Volume of rod $(V)$ is

$\begin{matrix} V = {πr}^{2} L \\ = π \times {(3.00 \times 10^{- 3})}^{2} \times 0.06 \\ = 1.69 \times 10^{- 6} m^{3} \end{matrix}$

Dipole moment $(μ)$ :

$\begin{matrix} μ = MV \\ = (2.70 \times 10^{3}) \times (1.69 \times 10^{- 6}) \\ = 4.5 \times 10^{- 3} {A.m}^{2} \end{matrix}$

Torque $(τ)$ :

$\begin{matrix} τ = μBsin (θ) \\ = (4.5 \times 10^{- 3}) \times (0.035) \times (\sin (68^{0})) \\ = 1.49 \times 10^{- 4} N . m \end{matrix}$

The magnitude of the torque on the rod due to $\bar{B}$ is $τ = 1.49 \times 10^{- 4} N . m$

(b) Calculations of the change in the orientation energy of the rod

$\begin{matrix} ΔU = - μB \times (\cos (θ_{f}) - \cos (θ_{i})) \\ = - 4.5 \times 10^{- 3} \times 0.035 \times (\cos (68) - \cos (34)) \\ = - 72.9 \times 10^{- 6} J \end{matrix}$