Chapter 31, Problem 31.13P

(a)

To determine

The magnetic flux through the loop due to the current.

Answer to Problem 31.13P

The magnetic flux through the loop due to the current is $\frac{{\mu }_{\text{0}}IL}{2\pi }\ln \left(1+\frac{w}{h}\right)$.

Explanation of Solution

Given info: The length of the rectangle is $1.00\text{m}$, the value of $b$ is $10.0\text{A}/\text{s}$, the value of $h$ is $1.00\text{cm}$ and the value of width of the rectangle is $10.0\text{cm}$.

The magnetic field is a function of the distance. So, the value of the magnetic field varies over the area of the rectangular loop.

Consider an infinitesimal section of the loop at a distance of $x$ from the wire of thickness $dx$.

Write the expression for the area of the infinitesimal section of the loop.

$dA=Ldx$

Here,

$L$ is the length of the rectangle.

$dx$ is the thickness of the infinitesimal section.

Write the expression for the magnetic field at the infinitesimal section.

$B=\frac{{\mu }_{\text{0}}I}{2\pi x}$

Here,

${\mu }_0$ is the vacuum permeability.

$x$ is the distance from the wire to the infinitesimal section.

$I$ is the current flowing in the wire.

Write the expression for the magnetic flux through the infinitesimal section.

$d\varphi =BdA$

Substitute $\frac{{\mu }_{\text{0}}I}{2\pi x}$ for $B$ and $Ldx$ for $dA$ in above equation.

$\begin{array}{c}d\varphi =\left(\frac{{\mu }_{\text{0}}I}{2\pi x}\right)\left(Ldx\right)\\ =\frac{{\mu }_{\text{0}}ILdx}{2\pi x}\end{array}$

Write the expression for the magnetic flux through the entire loop.

$\begin{array}{c}\varphi =\frac{{\mu }_{\text{0}}IL}{2\pi }{\displaystyle \underset{\text{h}}{\overset{\text{h}+\text{w}}{\int }}\frac{dx}{x}}\\ =\frac{{\mu }_{\text{0}}IL}{2\pi }{\left(\ln x\right)}_{\text{h}}^{\text{h}+\text{w}}\\ =\frac{{\mu }_{\text{0}}IL}{2\pi }\left(\ln \left(h+w\right)-\ln h\right)\\ =\frac{{\mu }_{\text{0}}IL}{2\pi }\ln \left(1+\frac{w}{h}\right)\end{array}$

Conclusion:

Therefore, the magnetic flux through the loop due to the current is $\frac{{\mu }_{\text{0}}IL}{2\pi }\ln \left(1+\frac{w}{h}\right)$.

(b)

To determine

The induced emf in the loop.

Answer to Problem 31.13P

The induced emf in the loop is $-4.80\text{μV}$.

Explanation of Solution

Given info: The length of the rectangle is $1.00\text{m}$, the value of $b$ is $10.0\text{A}/\text{s}$, the value of $h$ is $1.00\text{cm}$ and the value of width of the rectangle is $10.0\text{cm}$.

Write the expression for the induced emf in the loop.

$\epsilon =-\frac{d\varphi }{dt}$

Here,

$dt$ is the change in time.

From part (a), the magnetic flux through the loop due to the current is $\frac{{\mu }_{\text{0}}IL}{2\pi }\ln \left(1+\frac{w}{h}\right)$.

Substitute $\frac{{\mu }_{\text{0}}IL}{2\pi }\ln \left(1+\frac{w}{h}\right)$ for $\varphi$ in above equation.

$\begin{array}{c}\epsilon =-\frac{d\left(\frac{{\mu }_{\text{0}}IL}{2\pi }\ln \left(1+\frac{w}{h}\right)\right)}{dt}\\ =-\frac{{\mu }_{\text{0}}L}{2\pi }\ln \left(1+\frac{w}{h}\right)\frac{dI}{dt}\end{array}$

Substitute $a+bt$ for $I$ in the above equation.

$\begin{array}{c}\epsilon =-\frac{{\mu }_{\text{0}}L}{2\pi }\ln \left(1+\frac{w}{h}\right)\frac{d\left(a+bt\right)}{dt}\\ =-\frac{{\mu }_{\text{0}}L}{2\pi }\ln \left(1+\frac{w}{h}\right)\left(b\right)\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$, $1.00\text{m}$ for $L$, $10.0\text{cm}$ for $w$, $1.00\text{cm}$ for $h$ and $10.0\text{A}/\text{s}$ for $b$ in above equation.

$\begin{array}{c}\epsilon =-\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(1.00\text{m}\right)}{2\pi }\ln \left(1+\frac{10.0\text{cm}}{1.00\text{cm}}\right)\left(10.0\text{A}/\text{s}\right)\\ \approx -4.79\times {10}^{-6}\text{V}\times \frac{{10}^6\text{μV}}{1\text{V}}\\ \approx -4.80\text{μV}\end{array}$

Conclusion:

Therefore, the induced emf in the loop is $-4.80\text{μV}$.

(c)

To determine

The direction of the induced current in the rectangle.

Answer to Problem 31.13P

The direction of the induced current in the rectangle is counterclockwise.

Explanation of Solution

Given info: The length of the rectangle is $1.00\text{m}$, the value of $b$ is $10.0\text{A}/\text{s}$, the value of $h$ is $1.00\text{cm}$ and the value of width of the rectangle is $10.0\text{cm}$.

Write the expression for the current induced in the loop.

$I=\frac{\epsilon }{R}$

Here,

$R$ is the resistance.

From the above expression, the current is directly proportional to the emf induced in the loop. From part (b), the induced emf in the loop is $-4.80\text{μV}$. The value of the current is negative because the sign of the emf induction is negative which means the current flows in opposite direction from the direction which is shown in given figure. The negative sign indicates the counterclockwise direction by the right hand thumb rule. Thus, the current flows in the counterclockwise direction.

Conclusion:

Therefore, the direction of the induced current in the rectangle is counterclockwise.