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.