Question

Please solve and explain how its done.

Two identical circular thin-wire loops carry the same current I=7.8 A and are positioned "parallel to each other as illustrated in the figure below. The planes of the loops are perpendicular to the line connecting their centers 01 and 02. The radius of the loops a=0.9 cm and the distance between the loop planes b=1.53 cm. 0:0 G. What is the magnitude B1 of the magnetic field at point 01?: B1= What is the magnitude B2 of the magnetic field at point O2?: B2= What is the magnitude Bc of the magnetic field at point C in the middle between points O1 and O2?: Bce G. G.

Answer 1

Magnetic field at the point $O_1$ due to left coil is $B_{1L}=\frac{\mu_0 I}{2a}$

Magnetic field at the point $O_1$ due to right coil is $B_{1R}=\frac{\mu_0 Ia^2}{2(a^2+b^2)^{3/2}}$

Net magnetic field at point $O_1$ due to two coils is $B_{1}=\frac{\mu_0 I}{2a}+\frac{\mu_0 Ia^2}{2(a^2+b^2)^{3/2}}$

$B_{1}=\frac{4\pi*10^{-7}*7.8}{2*(0.9*10^{-2})}+\frac{4\pi*10^{-7}*7.8(0.9*10^{-2})^2}{2[(0.9*10^{-2})^2+(1.53*10^{-2})^2)^{3/2}]}$

$B_{1}=0.0005445+0.000071=0.000616\,T=6.16*10^{-4}\,T=6.16\,G$

----------------------

Magnetic field at the point $O_2$ due to left coil is $B_{2L}=\frac{\mu_0 Ia^2}{2(a^2+b^2)^{3/2}}$

Magnetic field at the point $O_2$ due to right coil is $B_{2R}=\frac{\mu_0 I}{2a}$

Net magnetic field at point $O_2$ due to two coils is $B_{2}=\frac{\mu_0 I}{2a}+\frac{\mu_0 Ia^2}{2(a^2+b^2)^{3/2}}$

$B_{2}=\frac{4\pi*10^{-7}*7.8}{2*(0.9*10^{-2})}+\frac{4\pi*10^{-7}*7.8(0.9*10^{-2})^2}{2[(0.9*10^{-2})^2+(1.53*10^{-2})^2)^{3/2}]}$

$B_{2}=0.0005445+0.000071=0.000616\,T=6.16*10^{-4}\,T=6.16\,G$

--------------------

Magnetic field at point C midway between the two coils is $B_{C}=B_{LC}+B_{RC}=\frac{\mu_0 Ia^2}{2(a^2+(b/2)^2)^{3/2}}+\frac{\mu_0 Ia^2}{2(a^2+(b/2)^2)^{3/2}}$

$B_{C}=\frac{2\mu_0 Ia^2}{2(a^2+(b/2)^2)^{3/2}}$

$B_{C}=\frac{\mu_0 Ia^2}{(a^2+(b/2)^2)^{3/2}}$

$B_C=\frac{4\pi*10^{-7}*7.8(0.9*10^{-2})^2}{[(0.9*10^{-2})^2+(1.53*10^{-2}/2)^2)^{3/2}]}$

$B_C=0.000142\,T=1.42*10^{-4}\,T=1.42\,G$

---------------------------