a) What is the magnitude of the total field at the point x = 0, z = 1.00 m in the xz-plane?
b) What is its direction?
Answer
The magnetic field due to stright wire is,
B=\left[\frac{\mu_{0}}{2 \pi} \frac{i}{r}\right]
here, \mu_{0} is the permebility of air or free space, i is the current and
r is the distance from conductor to imagined point \varphi is the arc angle.
a) From the above rule, the magnetic field at the point x=0 and z=1.0 \mathrm{~m} is,
B_{1} =-\frac{\mu_{0}}{2 \pi} \frac{(8.0 \mathrm{~A})}{(1.0 \mathrm{~m})}
=-\frac{\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)}{2 \pi} \frac{(8.0 \mathrm{~A})}{(1.0 \mathrm{~m})}
=-1.6 \times 10^{-6} \mathrm{~T}
Net magnetic field is,
B_{\text {ntt }} =1.5 \times 10^{-6} \mathrm{~T}-1.6 \times 10^{-6} \mathrm{~T}
=-10^{-7} \mathrm{~T}
b) Direction is along negative x axis.
c) From the above rule, the magnetic field at the point \mathrm{x}=1.0 \mathrm{~m} and z=0 is,
B_{1} =-\frac{\mu_{0}}{2 \pi} \frac{(8.0 \mathrm{~A})}{(1.0 \mathrm{~m})}
=-\frac{\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)}{2 \pi} \frac{(8.0 \mathrm{~A})}{(1.0 \mathrm{~m})}
=-1.6 \times 10^{-6} \mathrm{~T}
Net fiel d is,
\vec{B} =\vec{B}_{1}+\vec{B}_{0}
=\left(1.6 \times 10^{-6} \mathrm{~T}\right) \hat{k}+\left(1.5 \times 10^{-6} \mathrm{~T}\right) \hat{i}
Resultant field is,
B =\sqrt{\left(1.6 \times 10^{-6} \mathrm{~T}\right)^{2}+\left(1.5 \times 10^{-6} \mathrm{~T}\right)^{2}}
=2.19 \times 10^{-6} \mathrm{~T}
d) Drection of resultant is,
\theta =\tan ^{-1}\left(\frac{1.5 \times 10^{-6} \mathrm{~T}}{1.6 \times 10^{-6} \mathrm{~T}}\right)
=43.15^{\circ}
e) From the above rule, the magnetic field at the point \mathrm{x}=0 \mathrm{~m} and \mathrm{z}=-0.25 \mathrm{~m} is,
B_{1} =-\frac{\mu_{0}}{2 \pi} \frac{(8.0 \mathrm{~A})}{(0.25 \mathrm{~m})}
=-\frac{\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)}{2 \pi} \frac{(8.0 \mathrm{~A})}{(1.0 \mathrm{~m})}
=6.4 \times 10^{-6} \mathrm{~T}
Net magnetic field is,
B_{1}=\left(6.4 \times 10^{-6} \mathrm{~T}\right) \hat{i}+\left(1.5 \times 10^{-6} \mathrm{~T}\right) \hat{i}
=\left(7.9 \times 10^{-6} \mathrm{~T}\right) \hat{i}
f) Direction of resultant is along x axis.
