91影视

A=5.0cm

B=8.0cm

B=0.19T

I=6.2A

(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,

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

Put in equation (1),

$\mathrm{蟿}=\mathrm{ablB}$

ab=A

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

Put all the known values,

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

Thus, torque on the loop is$蟿=4.7脳{10}^{鈭�3}\mathrm{Nm}$

(b)for the magnetic moment of the loop

Magnetic moment is,

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

Thus, the magnetic moment of the loop is

(c)for maximum torque,

the maximum area when loop is circular,

2ab + 2b = 2$\mathrm{蟿蟿}$R

R is radius,

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

Area of the loop,

$A=蟺R^2=蟺{\left(\frac{a+b}{蟺}鈭�\right)}^2=\frac{(0.13m{)}^2}{蟺}=5.38脳{10}^{鈭�3}{\mathrm{m}}^2$

And the torque is,

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

Thus, maximum torque that can be obtained is$蟿=6.34脳{10}^{鈭�3}\mathrm{N}/\mathrm{m}$