91影视

The given data can be listed below as

• The magnetic field is, B = 3.00 T .
• The value of current is, I = 5.00A .

The term magnetic field may be defined as the area around the magnet behave like a magnet.

The right-angle triangle formed by the wire with

$\begin{array}{r}\mathrm{PQ}=0.600\mathrm{m},\mathrm{RP}=0.800\mathrm{m}\\ \mathrm{QR}=\sqrt{{\mathrm{PQ}}^2+{\mathrm{RP}}^2}\\ \mathrm{QR}=1.00\mathrm{m}\end{array}$

And the force act on each side$\stackrel{鈫�}{\mathrm{F}}=\stackrel{鈫�}{\mathrm{I}}脳\stackrel{鈫�}{\mathrm{B}}$ the angle between magnetic field and , magnetic field and , magnetic field and$\mathrm{RP}鈦�0^{鈭�},0.800,{90}^{鈭�}$

the force is calculated for three cases

$\begin{array}{r}{\mathrm{F}}_{\mathrm{PQ}}=\left(5\mathrm{A}\right)(0.600\mathrm{m})\left(3\mathrm{T}\right)\sin 鈦�\left(0^{鈭�}\right)\\ {\mathrm{F}}_{\mathrm{PQ}}=0\mathrm{N}\\ {\mathrm{F}}_{\mathrm{RP}}=\left(5\mathrm{A}\right)(0.800\mathrm{m})\left(3\mathrm{T}\right)\sin 鈦�\left({90}^{鈭�}\right)\\ {\mathrm{F}}_{\mathrm{PQ}}=12\mathrm{N}\end{array}$

$\begin{array}{r}{\mathrm{F}}_{\mathrm{RP}}=\left(5\mathrm{A}\right)(1.00\mathrm{m})\left(3\mathrm{T}\right)(0.800)\\ {\mathrm{F}}_{\mathrm{PQ}}=12\mathrm{N}\end{array}$

Hence, the force exerted by the magnetic field on each side of the triangle is ${\mathrm{F}}_{\mathrm{PQ}}=0\mathrm{N},{\mathrm{F}}_{\mathrm{RP}}=12\mathrm{N},{\mathrm{F}}_{\mathrm{QR}}=12\mathrm{N}$.

Now net force on the triangular loop of the wire is

$\begin{array}{r}\mathrm{F}={\mathrm{F}}_{\mathrm{QR}}鈭�{\mathrm{F}}_{\mathrm{RP}}\\ \mathrm{F}=12\mathrm{N}鈭�12\mathrm{N}\\ \mathrm{F}=0\end{array}$

Hence, the net force on the loop is zero.

Now the torque can be calculated as

For the force exerted is ${\mathrm{F}}_{\mathrm{PR}}$and the distance between the axis of rotation and wire is zero, for PQ force exerted is zero, and for QR the force exerted is${\mathrm{F}}_{\mathrm{QR}}$ And the distance between the center of the wire and the axis of the rotation is 0.300m.

Now calculated torque

$\begin{array}{r}{\mathrm{蟿}}_{\mathrm{PQ}}=\mathrm{r}\left(0\mathrm{N}\right)\\ {\mathrm{蟿}}_{\mathrm{PR}}=\left(0\mathrm{m}\right)\mathrm{Fsin}鈦�\left(\mathrm{f}\right)\\ {\mathrm{蟿}}_{\mathrm{PR}}=0\\ {\mathrm{蟿}}_{\mathrm{QR}}=(0.300\mathrm{m})\left(12\mathrm{N}\right)\sin 鈦�\left({90}^{鈭�}\right)\\ {\mathrm{蟿}}_{\mathrm{QR}}=3.60\mathrm{N}鈰�\mathrm{m}\end{array}$

the net torque is

$\begin{array}{r}\mathrm{蟿}={\mathrm{蟿}}_{\mathrm{PQ}}+{\mathrm{蟿}}_{\mathrm{RP}}+{\mathrm{蟿}}_{\mathrm{QR}}\\ \mathrm{蟿}=3.60\text{N.m}\end{array}$

Hence, the torque on each side of the loop is 3.60N.m

Now

The acting on the loop is calculated by the relation

$\begin{array}{r}\mathrm{蟿}=\mathrm{NIABsin}鈦�\left(\mathrm{f}\right)\\ \mathrm{蟿}=\left(1\right)\left(5\mathrm{A}\right)\left(\frac{1}{2}\right)(0.600\mathrm{m})(0.800\mathrm{m})(3.00\mathrm{T})\sin 鈦�\left({90}^{鈭�}\right)\\ \mathrm{蟿}=3.60\mathrm{N}鈰�\mathrm{m}\end{array}$

Hence, the magnitude of the net torque on the loop is 3.60 N

Now

Due to the force acting on the torque acting in the loop about the axis of which is out of plane so the net torque directed to rotate point Q out of the plane of the figure.