Chapter 30, Problem 5P

An aluminum ring of radius r₁ = 5.00 cm and resistance 3.00 × 10⁻⁴ Ω is placed around one end of a long air-core solenoid with 1 000 turns per meter and radius r₂ = 3.00 cm as shown in Figure P30.5. Assume the axial component of the field produced by the solenoid is one-half as strong over the area of the end of the solenoid as at the center of the solenoid. Also assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s. (a) What is the induced current in the ring? At the center of the ring, what are (b) the magnitude and (c) the direction of the magnetic field produced by the induced current in the ring?

Figure P30.5 Problems 5 and 6.

To determine

The induced current in the ring.

Answer to Problem 5P

The induced current in the ring is $1.6\text{A}$.

Explanation of Solution

Given info: Radius of aluminum ring is $5.00\text{cm}$, resistance of aluminum is $3.00\times {10}^{-4}\Omega$, number of turns per meter in solenoid is $1000\text{/m}$, radius at the other end is $3.00\text{cm}$ and current is increasing at a rate of $270\text{A}/\text{s}$.

The magnetic field due to solenoid can be given as,

$B={\mu }_{0}nI$

Here,

$B$ is the magnetic field due to solenoid.

${\mu }_0$ is the permeability of free space.

$n$ is the number of turns per meter in solenoid.

$I$ is the current in the coil.

The area of the coil can be given as,

$A=\pi r_2{}^2$

Here,

$A$ is the area of the coil.

$r_2$ is the radius of the coil.

Substitute $3.00\text{cm}$ for $r_2$ in the above equation to get $A$,

$\begin{array}{c}A=\pi {\left[(3.00\text{cm)}\left(\frac{1\text{m}}{100\text{cm}}\right)\right]}^2\\ =2.8274\times {\text{10}}^{-3}{\text{m}}^2\end{array}$

The emf generated in the coil can be given as,

$\begin{array}{c}\epsilon =-\frac{d{\Phi }_{\text{B}}}{dt}\\ =-\frac{d\left(BA\right)}{dt}\end{array}$

Here,

${\Phi }_{\text{B}}$ is the magnetic flux across the loop.

$\epsilon$ is the induced emf in the coil.

Substitute ${\mu }_{0}nI$ for $B$ in the above equation,

$\begin{array}{c}\epsilon =-\frac{d\left({\mu }_{0}nIA\right)}{dt}\\ =-{\mu }_{0}n\frac{dI}{dt}A\end{array}$

Substitute $1000\text{/m}$ for $n$, $4\pi \times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}$. for ${\mu }_0$, $2.8274\times {\text{10}}^{-3}{\text{m}}^2$ for $A$ and $270\text{A}/\text{s}$ for $\frac{dI}{dt}$ in the above equation,

$\begin{array}{c}\epsilon =-\left(4\pi \times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\left(1000\text{/m}\right)\left(270\text{A}/\text{s}\right)\left(2.8274\times {\text{10}}^{-3}{\text{m}}^2\right)\\ =9.6\times {10}^{-4}\text{V}\end{array}$

As the ring is placed around one end and also the field produced by the end of the solenoid is half at the centre of the solenoid.

Then, emf induced in the ring can be given as,

$\left|{\epsilon }_{\text{ring}}\right|=\left|\frac{\epsilon }{2}\right|$

Here,

${\epsilon }_{\text{ring}}$ is the emf induced in the ring.

Substitute $9.6\times {10}^{-4}\text{V}$ for $\epsilon$ in the above equation,

$\begin{array}{c}{\epsilon }_{\text{ring}}=\frac{9.6\times {10}^{-4}}{2}\text{V}\\ =4.8\times {10}^{-4}\text{V}\end{array}$

The current induced in the ring can be given as,

$I_{\text{ring}}=\frac{{\epsilon }_{\text{ring}}}{R}$

Here,

$I_{\text{ring}}$ is the current induced in the ring.

$R$ is the resistance of the ring.

Substitute $4.8\times {10}^{-4}\text{V}$ for ${\epsilon }_{\text{ring}}$ and $3.00\times {10}^{-4}\Omega$ for $R$ in the above equation,

$\begin{array}{c}{I}_{\text{ring}}=\frac{4.8\times {10}^{-4}\text{V}}{3.00\times {10}^{-4}\Omega }\\ =1.60\text{A}\end{array}$

Thus, the current induced in the ring is $1.60\text{A}$ in the counter clockwise direction when seen from the left side.

Conclusion:

Therefore, the current induced in the ring is $1.60\text{A}$ in the counter clockwise direction when seen from the left side.

To determine

The magnitude of the magnetic field at the centre of the ring.

Answer to Problem 5P

The magnitude of the magnetic field at the centre of the ring is $\text{20}\text{.1}\text{μT}$.

Explanation of Solution

Given info: Radius of aluminum ring is $5.00\text{cm}$, resistance of aluminum is $3.00\times {10}^{-4}\Omega$, number of turns per meter in solenoid is $1000\text{/m}$, radius at the other end is $3.00\text{cm}$ and current is increasing at a rate of $270\text{A}/\text{s}$.

The magnetic field at the center of the ring can be given as,

$B_{\text{ring}}=\frac{{\mu }_0{I}_{\text{ring}}}{2r_1}$

Here,

$B_{\text{ring}}$ is the magnetic field in the ring.

$r_1$ is the radius of aluminum ring.

Substitute $4\pi \times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}$ for ${\mu }_0$, $1.60\text{A}$ for $I_{\text{ring}}$ and $5.00\text{cm}$ for $r_1$ in the above equation,

$\begin{array}{c}{B}_{\text{ring}}=\frac{\left(4\pi \times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\left(1.60\text{A}\right)}{2\left(5.00\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =2.01\times {10}^{-5}\text{T}\\ =\text{20}\text{.1}\text{μT}\end{array}$

Thus, the magnitude of the magnetic field is $\text{20}\text{.1}\text{μT}$.

Conclusion:

Therefore, the magnitude of the magnetic field at the center of ring is $\text{20}\text{.1}\text{μT}$.

To determine

The direction of magnetic field at the center of the ring.

Answer to Problem 5P

The direction of magnetic field at the center of the ring is towards the left.

Explanation of Solution

Given info: Radius of aluminum ring is $5.00\text{cm}$, resistance of aluminum is $3.00\times {10}^{-4}\Omega$, number of turns per meter in solenoid is $1000\text{/m}$, radius at the other end is $3.00\text{cm}$ and current is increasing at a rate of $270\text{A}/\text{s}$.

Figure (1)

The solenoid’s field points to the right through the ring as shown in the figure (I). So, to oppose the increasing field, the direction of magnetic field at the center of the ring will be towards the left.

Conclusion:

Therefore, the direction of magnetic field at the center of the ring will be towards towards the left.