Question

In Figure a, two circular loops, with different currents but the same radius of $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$, are centered on a y axis. They are initially separated by distance $\mathit{L}\mathbf{=}\mathbf{\text{3.0\hspace{0.17em}cm}}$, with loop 2 positioned at the origin of the axis. The currents in the two loops produce a net magnetic field at the origin, with y component ${\mathit{B}}_{\mathbf{y}}$. That component is to be measured as loop 2 is gradually moved in the positive direction of the y axis. Figure b gives ${\mathit{B}}_{\mathbf{y}}$as a function of the position $\mathit{y}$of loop 2. The curve approaches an asymptote of ${\mathit{B}}_{\mathbf{y}}\mathbf{=}\mathbf{7}\mathbf{.20}\mathbf{\text{\hspace{0.17em}μT}}$as $\mathit{y}\mathbf{\to }\mathbf{\infty }$. The horizontal scale is set by${\mathit{y}}_{\mathbf{s}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$. (a) What are current ${\mathit{i}}_{\mathbf{1}}$in loop 1 and (b) What are current ${\mathit{i}}_{\mathbf{2}}$ in loop 2?

Short answer

(a) The current $i_1$in the loop 1 is $0.90\text{A}$.

(b) The current $i_2$in the loop 2 is $2.7\text{A}$.

Step-by-step solution

Identification of given data:

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 $y\to \infty$, $B_y=7.20\text{μT}$

Significance of magnetic field:

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.

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

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

Explanation:

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

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

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,

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

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

The equation for magnetic field can be written as,

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

(a) Determining the current i1 in the loop 1:

As $y\to \infty$,the equation of magnetic field becomes,

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

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.}}}{{\mu }_0{R}^2}$

Substitute known values in the above equation.

role="math" localid="1663242054791" $\begin{array}{rcl}{i}_1& =& \frac{2\times 7.20\times {10}^{-6}\text{\hspace{0.17em}T}\times {({(0.04\text{m})}^2+{(0.03\text{\hspace{0.17em}m})}^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{A}}^2}$}\right.}\times {\left(0.04\text{\hspace{0.17em}m}\right)}^2}\\ & =& \frac{14.4\times {10}^{-6}\text{\hspace{0.17em}T}\times 1.25\times {10}^{-4}{\text{\hspace{0.17em}m}}^3}{4\times 3.14\times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}\times 1.6\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{2}}}\\ & =& 0.8952\text{A}\\ & =& 0.90\text{A}\end{array}$

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

(b) Determining the current i2 in the loop 2:

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),

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

By substituting the values, you get

$\frac{0.90\text{A}}{{({\left(0.04\text{\hspace{0.17em}m}\right)}^2+{\left(0.03\text{\hspace{0.17em}m}\right)}^2)}^{\frac{3}{2}}}=\frac{i_2}{{({\left(0.04\text{\hspace{0.17em}m}\right)}^2+{\left(0.06\text{\hspace{0.17em}m}\right)}^2)}^{\frac{3}{2}}}$

$\begin{array}{rcl}{i}_2& =& \frac{0.90\text{\hspace{0.17em}A}\times 3.75\times {10}^{-4}{\text{m}}^3}{1.25\times {10}^{-4}{\text{m}}^3}\\ & =& 2.7\text{A}\end{array}$

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