Question

the figure below shows loops of wire moving to the right with equal velocity and exiting the same region of constant magnetic field. assume that there are only two lengths appearing in the figure l and 2l, and that the resistance per unit length of the wires is identical. rank the figures in terms of the current generated in the loops. (1) (2) (3) (4) o2=3>1=4 o1=2>3=4 o1>2>3>4 o1>2=4>3

Explanation:

Step1: Calculate the induced emf

The induced emf \(\mathcal{E}=BLv\), where \(B\) is the magnetic field, \(L\) is the length of the wire in the magnetic field, and \(v\) is the velocity.

Step2: Calculate the resistance

Let the resistance per unit length be \(r\). The resistance \(R = r\times\) (perimeter of the loop).
For loop (1): perimeter \(P_1= 2(L + 2L)=6L\), \(R_1 = 6Lr\), \(\mathcal{E}_1=BLv\), \(I_1=\frac{\mathcal{E}_1}{R_1}=\frac{BLv}{6Lr}=\frac{vB}{6r}\)
For loop (2): perimeter \(P_2 = 4\times2L = 8L\), \(R_2=8Lr\), \(\mathcal{E}_2 = 2Lv\), \(I_2=\frac{\mathcal{E}_2}{R_2}=\frac{2LvB}{8Lr}=\frac{vB}{4r}\)
For loop (3): perimeter \(P_3=2(2L + L)=6L\), \(\mathcal{E}_3 = 2Lv\), \(R_3 = 6Lr\), \(I_3=\frac{\mathcal{E}_3}{R_3}=\frac{2LvB}{6Lr}=\frac{vB}{3r}\)
For loop (4): perimeter \(P_4=4L\), \(R_4 = 4Lr\), \(\mathcal{E}_4=LvB\), \(I_4=\frac{\mathcal{E}_4}{R_4}=\frac{LvB}{4Lr}=\frac{vB}{4r}\)

Comparing the currents: \(I_3>I_2 = I_4>I_1\)

Answer:

\(1<2 = 4<3\) (Since the options are in terms of ranks, and from our calculation of currents \(I_3>I_2 = I_4>I_1\), the rank order is \(1<2 = 4<3\))