Figures 30-32 give four situations in which we pull rectangular wire loops out of identical magnetic fields page) at the same constant speed. The loops have edge lengths of either L or 2L, as drawn. Rank the situations according to (a) the magnitude of the force required of us and (b) the rate at which energy is transferred from us to the thermal energy of the loop greatest first.

i) Fig.30-32.

ii) The magnetic field is identical in all given loops.

iii) The speed v is constant in all given loops.

iv) The loops have an edge length of either L or 2L .

Using Eq.30-2, and Eq.30-4, find the corresponding magnitude of the induced emf. Using this value of emf in Ohm’s law equation, find the corresponding current. Finally, by using the formula for the force, find the corresponding values of force required to pull out the loop. Comparing these values of forces, find the ranks of the situations according to the magnitude of the force required of us to pull out of the loop. Also, by using the formula for the rate of transfer of energy from us to the thermal energy, find the ranks of the situations.

Formulae are as follows:

${\Phi }_B=BA$

$\left|\epsilon \right|=\frac{d{\Phi }_B}{dt}$

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

$F=iLB$

Rate of transfered = F x v

Let us consider that the loop has a length inside the magnetic field. Now, from Eq.30-2, the magnetic field through the loop is given by,

${\Phi }_B=BA......................................................................\left(30-2\right)$

Where, B is the magnetic field and A is the area of the loop, and is,

A = Lx

Therefore,

${\Phi }_B=BLx$

According to Faraday’s law, the magnitude of induced emf is given by,

$\left|\epsilon \right|=\frac{d{\Phi }_B}{dt}................................................................................\left(30-4\right)$

Putting the value on, it gives,

$$\begin{gathered} \left|\epsilon \right|=\frac{d\left(BLx\right)}{dt} \\ \left|\epsilon \right|=BL\frac{dx}{dt} \end{gathered}$$

But, $\frac{dx}{dt}=v$

$\left|\epsilon \right|=BLv.....................................................................\left(1\right)$

The resistance of the wire in the loop is directly proportional to the total length of the loop. Because the material is the same in all the given loops. Let us consider k the proportionality constant.

According to Ohm’s law, the current is given by,

$i=\frac{\epsilon }{R}...........................................................\left(2\right)$

And the resistive force opposing the motion of the loop would be,

$F=iLB..........................................................................\left(3\right)$

And therefore,

In loop 1

From Eq.1, the emf is,

$\left|\epsilon \right|=B\left(2L\right)v$

And from Fig.30-32, the resistance is given by,

$R=6Lk$

Where, k is proportionality constant.

Therefore, from Eq.2, it gives,

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ i=\frac{2BLk}{6Lk} \end{gathered}$$

Therefore, from Eq.3, the force $F_1$required to pull out the loop is given by,

$F_1=\frac{2B^2{L}^{2}v}{3Lk}$

Similarly, in loop 2,

From Eq.1, the emf is,

$\left|\epsilon \right|=B\left(2L\right)v$

And from Fig.30-32, the resistance is given by,

$R=8Lk$

Where, k is the proportionality constant.

Therefore, from Eq.2 it gives,

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ i=\frac{2BLv}{8Lk} \end{gathered}$$

Therefore, from Eq.3, the force $F_2$required to pull out the loop is given by,

$F_2=\frac{B^2{L}^{2}v}{2Lk}$

And in loop 3

From Eq.1, the emf is,

$\left|\epsilon \right|=B\left(L\right)v$

And from Fig.30-32, the resistance is given by,

R = 6Lk

Where, is the proportionality constant.

Therefore, from Eq.2, it gives,

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ i=\frac{BLv}{6Lk} \end{gathered}$$

Therefore, from Eq.3, the force $F_3$required to pull out the loop is given by,

$F_3=\frac{B^2{L}^{2}v}{6Lk}$

Also, in loop 4,

From Eq.1, the emf is,

$\left|\epsilon \right|=B\left(L\right)v$

And Lk

Where, k is proportionality constant.

Therefore, from Eq.2, it gives,

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ i=\frac{BLv}{4Lk}{F}_4 \end{gathered}$$

$F_4=\frac{B^2{L}^{2}v}{4Lk}$Therefore, from Eq.3, the force required to pull out the loop is given by,

Comparing these values of force requires pulling out the loop, it gives,

$F_1>F_2>F_4>F_3$

Therefore, rank the situations as 1) Loop 1, 2) Loop 2, 3) Loop 4, and 4) Loop 3.

The rate of energy from us to the thermal energy is given by,

Rate 0f energy transferred = F x v

Since, velocity v is the same for all the given loops. So, the rate of energy transferred is directly dependent on the force F required to pull out of the loop. But, from the solution a) ,

$F_1>F_2>F_4>F_3$

Therefore, the ranks of the loops according to the rate of the energy transferred from us to the thermal energy of the loop as 1) Loop 1, 2) Loop 2, 3) Loop 4, and 4) Loop 3.

Using Faraday’s law and Ohm’s law, and also using the formula for the force required to pull out the loop and the rate of energy from us to the thermal energy, answer this question.