The $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{cm}\mathbf{\times }\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{cm}$rectangular circuit shown in Fig. E27.41 is hinged alongside ab. It carries a clockwise 5.00 A current and is located in a uniform 1.20 T magnetic field oriented perpendicular to two of its sides, as shown. (a) Draw a clear diagram showing the direction of the force that the magnetic field exerts on each segment of the circuit (ab, bc, etc.). (b) Of the four forces you drew in part (a), decide which ones exert a torque about the hinge ab. Then calculate only those forces that exert this torque. (c) Use your results from part (b) to calculate the torque that the magnetic field exerts on the circuit about the hinge axis ab.

It is a unit of measurement for the amount of force that may cause an item to rotate around an axis.

Rectangular circuit has, ad = cd = 20.0 cm

Rectangular circuit has, bc = ad = 35.0 cm

The current is, l = 5.00A = 5.00 A

The uniform magnetic field is, B = 1.20 T

(a)

Current is in upward direction in ab and downward in cd while magnetic field is in right side.by right hand rule direction of force acting on ab inwards and cd is outwards.no force acts on bc and ad. thus, current is parallel and anti-parallel to magnetic field.

(b)

Force acting on ab will not exert any torque because circuit is hinged alongside ab. force acting on cd tends to rotate loop about hinge and thus it produces torque about axis. Force acting on wire is,

$F=llB\sin \varphi$

Here I is the current, l is the wire length and B is the magnetic field.

Substitute all the value in the above equation.

$\begin{array}{r}{F}_{cd}=IIB\sin \left(\varphi \right)\\ =\left(5\mathrm{A}\right)\times \left(0.200\mathrm{m}\right)\times \left(1.20\mathrm{T}\right)\times \sin \left(90\right)\\ =1.20\mathrm{N}\end{array}$

Thus, the forces that exert this torque is${\mathrm{F}}_{\mathrm{cd}}=1.20\mathrm{N}$

(c)

The torque is expressed as,

$\begin{array}{r}\tau =F_{cd}r\\ =\left(1.20\mathrm{N}\right)\left(0.350\mathrm{m}\right)\\ =0.420\mathrm{Nm}\\ \tau =0.420\mathrm{Nm}\end{array}$

Thus, the torque that the magnetic field exerts on the circuit about the hinge axis ab is$\tau =0.420\mathrm{Nm}$ .