91Ó°ÊÓ

1. The current i₁ in the larger coil 1 is fixed.
2. The current i₂ in the coil 2 can be varied.
3. The vertical axis is set by${\mathrm{μ}}_{(net,s)}=2.0×{10}^{-5}A.m^2$.
4. The horizontal axis is set by i₂s=10.0mA.
5. The current in the coil 2 is reversed.
6. The current i₂=7.0mA.

The orientation and magnetic strength of a magnet or other item that generates a magnetic field are referred to as magnetic moment.

By using the equation 28-35, we can find the product N₂A₂when i₂=0.0050A. Using this value ofproduct $N_2{A}_2$and the magnetic momentrole="math" localid="1662974914653" ${\mathrm{μ}}_{net,s}$, we can find the total magnetic moment of the two-coil system when i₂=7.0mA.

Formula

1. From equation 28-35, the magnetic moment is$\mathrm{μ}=NAi$
2. The total moment is$\mathrm{μ}={\mathrm{μ}}_{(}net,s)+N_2{A}_2{i}_2$

From the graph, consider the point corresponding to i₂=0. This means that coil has no magnetic moment. Hence, the magnetic moment of coil 1 must be

${\mathrm{μ}}_1=2.0×{10}^{-5}A.m^2$

When i₂,=0.0050A the magnetic moment of coil 2 will be

.${\mathrm{μ}}_2=2.0×{10}^{-5}A.m^2$

Therefore. From equation 28-35, we get

$$\begin{gathered} N_2{A}_2={\mathrm{μ}}_2/i_2 \\ {N}_2{A}_2=(2.0×{10}^{-5}A.m^2)/0.0050A \end{gathered}$$

$N_2{A}_2=4.0×{10}^{-3}{m}^2$

Now since the direction of the current of the second coil is changed, the net moment is the sum of two positive contributions from coil 1 and coil 2.

For, i₂=0.007A the total moment is

$\mathrm{μ}={\mathrm{μ}}_{net,s}+N_2{A}_2{i}_2$

$\mathrm{μ}=2.0×{10}^{-5}A.m^2+\left[\right(4.0×{10}^{-3}{m}^2\left)\right(0.007A\left)\right]$

$\mathrm{μ}=4.8×{10}^{-5}A.m^2$