Question

The plane of a 5.0 cm * 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

Short answer

(a)torque on the loop is$\tau =4.7\times {10}^{-3}\mathrm{Nm}$

(b) the magnetic moment of the loop is$\mu =0.025{\mathrm{Am}}^2$

(c) maximum torque that can be obtained is$\tau =6.34\times {10}^{-3}\mathrm{N}/\mathrm{m}$

Step-by-step solution

Step 1:

A=5.0cm

B=8.0cm

B=0.19T

I=6.2A

Step 2:

(a)for torque on the loop

Force acting is in two parallel wires in loop.one force in and another out but bath has same magnitude,

$\begin{array}{r}\tau =Fr\\ \mathrm{F}=\mathrm{laB}\end{array}$ (1)

Put in equation (1),

$\mathrm{\tau }=\mathrm{ablB}$

ab=A

role="math" localid="1668234520623" $\mathrm{\tau }=A\mathrm{lB}$

Put all the known values,

$\begin{array}{r}\mathrm{T}=\left(6.0\mathrm{A}\right)\left(0.050\mathrm{m}\right)\left(0.080\mathrm{m}\right)\left(0.19\mathrm{T}\right)=4.7\times {10}^{-3}\mathrm{Nm}\\ \mathrm{T}=4.7\times {10}^{-3}\mathrm{Nm}\end{array}$

Thus, torque on the loop is$\tau =4.7\times {10}^{-3}\mathrm{Nm}$

Step (3)

(b)for the magnetic moment of the loop

Magnetic moment is,

$\begin{array}{r}\mathrm{\mu }=\mathrm{IA}\\ \mathrm{\mu }=(6.2\mathrm{A})(0.050\mathrm{m})(0.080\mathrm{m})=0.025{\mathrm{Am}}^2\\ \mathrm{\mu }=0.025{\mathrm{Am}}^2\end{array}$

Thus, the magnetic moment of the loop is

Step 4:

(c)for maximum torque,

the maximum area when loop is circular,

2ab + 2b = 2$\mathrm{\tau \tau }$R

R is radius,

$\mathrm{R}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{\tau \tau }}$

Area of the loop,

$A=\pi R^2=\pi {\left(\frac{a+b}{\pi }-\right)}^2=\frac{(0.13m{)}^2}{\pi }=5.38\times {10}^{-3}{\mathrm{m}}^2$

And the torque is,

$\begin{array}{r}T=\left(6.2\mathrm{A}\right)\left(5.38\times {10}^{-3}{\mathrm{m}}^2\right)\left(0.19\mathrm{T}\right)=6.34\times {10}^{-3}\mathrm{N}/\mathrm{m}\\ T=6.34\times {10}^{-3}\mathrm{N}/\mathrm{m}\end{array}$

Thus, maximum torque that can be obtained is$\tau =6.34\times {10}^{-3}\mathrm{N}/\mathrm{m}$