91Ó°ÊÓ

The radius, $R_1=R_2=4\text{cm}=0.04\text{m}$

The length, $L=3\text{cm}=0.03\text{m}$

The horizontal scale distance, $y_s=10\text{cm}=0.01\text{m}$

The magnetic field with$y$ component as $y→∞$, $B_y=7.20\text{μ°Õ}$

The area in which the force of magnetism acts around a magnetic material or a moving electric charge is known as the magnetic field.

You can find the net magnetic field along the y direction due to both the coils. By using the given conditions in the problem and the graph, you can find the currents in both the loops.

Formula:

The magnetic field is defined by using following formula.

$B=\frac{{\mathrm{μ}}_{0}i{R}^2}{2{\left(R^2+Z^2\right)}^{\frac{3}{2}}}$

Here, $i$is the current,${\mathrm{μ}}_0$ is the permeability of free space having a value $4\mathrm{π}×{10}^{−7}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^{\text{2}}$}\right.$,$R$ is the radius,$Z$ and is the distance.

You know the magnetic field due to a coil at a distance $z$is,

$B=\frac{{\mathrm{μ}}_{0}i{R}^2}{2{\left(R^2+z^2\right)}^{\frac{3}{2}}}$

The net magnetic field due to both the coils is,

$B_y=B_1+B_2$

But from the graph, you can conclude that the magnetic field due to loop is negative. Therefore,

$B_y=\frac{{\mathrm{μ}}_0{i}_1{R}^2}{2{\left(R^2+z_1^2\right)}^{\frac{3}{2}}}−\frac{{\mathrm{μ}}_0{i}_2{R}^2}{2{\left(R^2+z_2^2\right)}^{\frac{3}{2}}}$

As $z_1^2=L^2$and $z_2^2=y^2$.

The equation for magnetic field can be written as,

$B_y=\frac{{\mathrm{μ}}_0{i}_1{R}^2}{2{\left(R^2+L^2\right)}^{\frac{3}{2}}}−\frac{{\mathrm{μ}}_0{i}_2{R}^2}{2{\left(R^2+y^2\right)}^{\frac{3}{2}}}$ ..... (i)

As $y→∞$,the equation of magnetic field becomes,

$B_y=\frac{{\mathrm{μ}}_0{i}_1{R}^2}{2{\left(R^2+L^2\right)}^{\frac{3}{2}}}$

Rearranging the above equation for current $i_1$.

role="math" localid="1663242063902" $i_1=\frac{2B_y{(R^2+L^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\mathrm{μ}}_0{R}^2}$

Substitute known values in the above equation.

role="math" localid="1663242054791" $\begin{array}{rcl}{i}_1& =& \frac{2×7.20×{10}^{−6}\text{â€Í¿}×{({(0.04\text{m})}^2+{(0.03\text{″¾})}^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{4\mathrm{Ï€}×{10}^{−7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{A}}^2}$}\right.}×{\left(0.04\text{″¾}\right)}^2}\\ & =& \frac{14.4×{10}^{−6}\text{â€Í¿}×1.25×{10}^{−4}{\text{″¾}}^3}{4×3.14×{10}^{−7}\text{N}/{\text{A}}^{\text{2}}×1.6×{10}^{−3}{\text{″¾}}^{\text{2}}}\\ & =& 0.8952\text{A}\\ & =& 0.90\text{A}\end{array}$

Hence, the current $i_1$in the loop 1 is$0.90\text{A}$.

Now from the graph, the horizontal scale is set as,

$y_s=10\text{cm}=0.01\text{m}$

Therefore, you can say that $B_y=0$at $y=0.06\text{m}$.

From equation (i),

$$\begin{gathered} −\frac{{\mathrm{μ}}_0{i}_2{R}^2}{2{\left(R^2+y^2\right)}^{\frac{3}{2}}}=0 \\ \frac{{\mathrm{μ}}_0{i}_1{R}^2}{2{\left(R^2+L^2\right)}^{\frac{3}{2}}}=\frac{{\mathrm{μ}}_0{i}_2{R}^2}{2{\left(R^2+y^2\right)}^{\frac{3}{2}}} \\ \frac{i_1}{{\left(R^2+L^2\right)}^{\frac{3}{2}}}=\frac{i_2}{{\left(R^2+y^2\right)}^{\frac{3}{2}}} \end{gathered}$$

By substituting the values, you get

$\frac{0.90\text{A}}{{({\left(0.04\text{″¾}\right)}^2+{\left(0.03\text{″¾}\right)}^2)}^{\frac{3}{2}}}=\frac{i_2}{{({\left(0.04\text{″¾}\right)}^2+{\left(0.06\text{″¾}\right)}^2)}^{\frac{3}{2}}}$

$\begin{array}{rcl}{i}_2& =& \frac{0.90\text{‼î}×3.75×{10}^{−4}{\text{m}}^3}{1.25×{10}^{−4}{\text{m}}^3}\\ & =& 2.7\text{A}\end{array}$

Hence, the current $i_2$in the loop 2 isrole="math" localid="1663242627340" $2.7\text{A}$.