Question

In the discussion of motional emf (Sect. 7.1.3) $I$assumed that the wire loop (Fig. 7.10) has a resistance $R$; the current generated is then $I=\frac{vBh}{R}$. But what if the wire is made out of perfectly conducting material, so that $R$is zero? In that case, the current is limited only by the back emf associated with the self-inductance$L$ of the loop (which would ordinarily be negligible in comparison with $IR$). Show that in this regime the loop (mass$m$ ) executes simple harmonic motion, and find its frequency. [Answer: $\omega =Bh/\sqrt{mL}$].

Short answer

The expression for the frequency is $\frac{Bh}{\sqrt{mL}}$.

Step-by-step solution

Write the given data from the question.

The length of the loop is $l$.

The resistance of the loop is $R$.

The self-inductance is $L$.

The mass of the loop is $m$.

The generated current in the loop is $I=\frac{vBh}{R}$.

Determine the formulas to calculate the frequency.

The expression to calculate the back emf is given as follows.

$\epsilon =Bhv$ ……. (1)

The expression to calculate the back emf associated with the self-inductance is given as follows.

$\epsilon =-L\frac{dI}{dt}$ ……. (2)

The expression for the force on the wire is given as follows.

$F=Blh$ …….. (3)

The expression for the force on the wire in terms of mass is given as follows.

$F=m\frac{dv}{dt}$ …….. (4)

Calculate the frequency of the simple harmonic.

Equate the forces on the wire.

From the equation (3) and (4).

$\begin{array}{c}m\frac{dv}{dt}=Blh\\ \frac{dv}{dt}=\frac{Blh}{m}\end{array}$

Differentiate the above equation with respect to $t$.

$\frac{d^{2}v}{dt}=\frac{Bl}{m}\frac{dI}{dt}$ ……… (5)

Equate the equation (1) and (2).

$\begin{array}{l}-L\frac{dI}{dt}=Bhv\\ \frac{dI}{dt}=-\frac{Bhv}{L}\end{array}$

Substitute $-\frac{Bhv}{L}$for $\frac{dI}{dt}$into equation (5).

$\begin{array}{l}\frac{d^{2}v}{dt}=\frac{Bl}{m}(-\frac{Bhv}{L})\\ \frac{d^{2}v}{dt}=-\frac{B^{2}lhv}{mL}\end{array}$

From the above equation,

$\begin{array}{l}{\omega }^2=\frac{B^2{h}^2}{mL}\\ \omega =\sqrt{\frac{B^2{h}^2}{mL}}\\ \omega =\frac{Bh}{\sqrt{mL}}\end{array}$

Hence the expression for the frequency is $\frac{Bh}{\sqrt{mL}}$.