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. Current $i_{2}$ in coil $2$ can be varied. Figure (b) gives the net magnetic moment of the two-coil system as a function of $i_{2}$ . The vertical axis scale is set by, $μ_{nets} = 2.0 \times 10^{- 5} A m^{2}$ and the horizontal axis scale is set by, $i_{2 s} = 20.0 mA$ . If the current in coil $2$ is then reversed, what is the magnitude of the net magnetic moment of the two-coil system when $i_{2} = 7.0 mA ?$

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$(b)$ Question Image

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Accepted Answer

Given:

The current in the larger coil $1$ , which is fixed, is $i_{1}$ .

Current in the smaller coil $2$ , which can be varied, is $i_{2}$ .

In the graph, the vertical axis scale is set by, $μ_{nets} = 2.0 \times 10^{- 5} A m^{2}$ .

The horizontal axis scale is set by $i_{2 s} = 20.0 mA$ .

The magnetic moment of current-carrying coil is given by,

$μ = N i A$ ...(1)

where, $N$ is the number of turns, $i$ is the current in the coil and $A$ is the area of coil.

From the figure (b),

when, $i_{2} = 0$ , then the magnitude of magnetic moment of coil $2$ ,

$\Rightarrow μ_{2} = 0$ .

The net magnetic moment is given by,

$\Rightarrow μ_{nets} = μ_{1} + μ_{2}$ ...(2)

$\Rightarrow μ_{1} = 2.0 \times 10^{- 5} A m^{2}$

Now, looking at the point where the line crosses the axis at $i_{2} = 10.0 mA$ , net magnetic moment is zero.

$\Rightarrow μ_{nets} = μ_{1} + μ_{2} = 0$ ...(3)

$μ_{2} = - μ_{1} = - 2.0 \times 10^{- 5} A m^{2}$ .

When, $i_{2} = 10 mA = 10 \times 10^{- 3} m$ , using equation (1), we get,

$N_{2} A_{2} = \frac{μ_{2}}{i_{2}}$

$\Rightarrow N_{2} A_{2} = \frac{2.0 \times 10^{- 5}}{10 \times 10^{- 3}}$

$\Rightarrow N_{2} A_{2} = 2.0 \times 10^{- 3} m^{2}$ .

Now, the direction of current in the coil $2$ has changed so that the net moment is the sum of two (positive) contributions from coil $1$ and coil $2$ when $i_{2} = 7 mA$ .

Then, the total magnetic moment is given by,

$μ_{net} = μ_{1} + μ_{2} = (2.0 \times 10^{- 5}) + (N_{2} A_{2} i_{2})$

$\Rightarrow μ_{net} = (2.0 \times 10^{- 5}) + (2.0 \times 10^{- 3} \times 7 \times 10^{- 3})$

$\Rightarrow μ_{net} = 3.4 \times 10^{- 5} A m^{2}$ .

Resnick & Halliday Solutions for Chapter: Magnetic Fields, Exercise 1: Problems

Resnick & Halliday Physics Solutions for Exercise - Resnick & Halliday Solutions for Chapter: Magnetic Fields, Exercise 1: Problems

Questions from Resnick & Halliday Solutions for Chapter: Magnetic Fields, Exercise 1: Problems with Hints & Solutions

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