Question

In (Figure (a)), two concentric coils, lying in the same plane, carry currents in opposite directions. The current in the larger coil 1 is fixed. Currentin coil 2 can be varied. (Figure (b))gives the net magnetic moment of the two-coil system as a function of i₂. The vertical axis scale is set by ${\mathbf{\mu }}_{\mathbf{(}\mathbf{net}\mathbf{,}\mathbf{x}\mathbf{)}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{A}\mathbf{\cdot }$and the horizontal axis scale setby i_2x=10.0mA. If the current in coil 2 is then reversed, what is the magnitude of the net magnetic moment of the two-coil system when i₂=7.0mA?

Short answer

The magnitude of net magnetic moment of the two-coil system when i₂=7.0mA is $\mu =4.8\times {10}^{-5}A.m^2.$

Step-by-step solution

Identification of given data

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${\mu }_{(net,s)}=2.0\times {10}^{-5}A.m^2$.
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.

Significance of magnetic moment

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" ${\mu }_{net,s}$, 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$\mu =NAi$
2. The total moment is$\mu ={\mu }_{(}net,s)+N_2{A}_2{i}_2$

Determining the magnitude of net magnetic moment of the two-coil system when i2=7.0mA.

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

${\mu }_1=2.0\times {10}^{-5}A.m^2$

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

.${\mu }_2=2.0\times {10}^{-5}A.m^2$

Therefore. From equation 28-35, we get

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

$N_2{A}_2=4.0\times {10}^{-3}{m}^2$

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

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

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

$\mu =4.8\times {10}^{-5}A.m^2$