Question

FIGURE 4.1 shows a rectangular wire loop 0.3 m x 0.2 m moving horizontally to the right at 12 ms -1 in a uniform magnetic field of 0.8 T. The induced current in the wire is 3 A.
(i) Determine the resistance of the wire loop
(ii) Determine the direction of the induced current. Explain how you determine the direction of the induced current.​

Answer 1

Given :-

• Length of the rectangular wire, L=0.3 m
• Width of the rectangular wire, b=0.2m
• Magnetic field strength, B=0.8 T
• Velocity of the loop, v =12 m/s
• Induced Current, I = 3 A

$\underline{\underline{\large\bf{Solution:-}}}$

(I) Emf developed,E in the loop is given as:

$\begin{gathered}\\\implies\quad \sf E = BLv \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf E = 0.8 \times 0.3 \times 12 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf E = 2.88 V \\\end{gathered}$

$\longrightarrow$I = E/R

$\quad$R = E/I

where

• R = resistance
• E = Induced EMF
• I = Current

$\begin{gathered}\\\implies\quad \sf R = \frac{2.88}{3} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \boxed{\sf{ R = 0.96 \;ohm}}\\\end{gathered}$

(ii) The direction of current induced is from P to Q which is given by B × V vector . It may also be explained by Lenz law. Since magnetic field is from S to N . The fingers of the right hand are placed around the wire so that the curling of fingers will show the direction of the magnetic field produced by the wire then the thumb points in the direction of current flow which is from P to Q.