Answer:
[tex]6.28\times 10^{-5}\ \text{T}[/tex]
[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]
[tex]I_l[/tex] = Current in circular loop = 13 A
[tex]r_l[/tex] = Radius of circular loop = 13 cm
[tex]N[/tex] = Number of turns = 58
[tex]r_c[/tex] = Radius of coil = 0.94 cm
[tex]I_c[/tex] = Current in coil = 1.9 A
[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]
Magnitude of magnetic field in circular loop
[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]
The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].
Torque is given by
[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]
The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].
