Answer:
[tex]5.39\cdot 10^{-3} N\cdot m[/tex]
Explanation:
The torque exerted on a current single loop of wires is given by:
[tex]\tau = \mu B sin \theta[/tex]
where
[tex]\mu[/tex] is the magnetic moment of the loop
B is the strength of the magnetic field
[tex]\theta[/tex] is the angle between the direction of the field and the normal to the plane of the loop
The magnetic moment of a loop is given by
[tex]\mu =IA[/tex]
where
I is the current in the loop
A is the area enclosed by the coil
Substituting into the original equation,
[tex]\tau=IABsin \theta[/tex]
in this problem, we have:
I = 20.0 mA = 0.020 A is the current in the loop
[tex]c=2.15 m[/tex] is the circumference of the loop, so we can find its radius:
[tex]c=2\pi r\\r=\frac{c}{2\pi}=\frac{2.15}{2\pi}=0.342 m[/tex]
So the area of the coil is
[tex]A=\pi r^2=\pi(0.342)^2=0.367 m^2[/tex]
Also we have
B = 0.735 T (strength of the magnetic field)
[tex]\theta=90^{\circ}[/tex] (because the field is parallel to the plane of the coil)
So, the magnitude of the torque is:
[tex]\tau=(0.020)(0.367)(0.735)(sin 90^{\circ})=5.39\cdot 10^{-3} N\cdot m[/tex]
