When a conducting loop of resistance `10 Omega` and area 10 `cm^(2)` is removed from an external magnetic field acting normally, the variation of induced current in the loop with time is shown in the figure.

Find the
(i) total charge passed through the loop.
(ii) change in magnetic flux through the loop.
(iii) magnitude of the magnetic field applied.

Explanation

(i) Total charge pased through the loop q=Area of l-t graph `= (1)/(2)xx(1.0s) xx(0.4) A=0.2 C`
(ii) Since |q| = `("Change in magnetic flux" Deltaphi_(B))/("Resistance of loop R") ` and R = 10 `Omega`
`implies Delta phi_(B) = Rxxq = 10xx0.2 = 2.0 Wb`
(iii) Since `Delta phi_(B)= phi_(1)-phi_(2)` and `phi_(2)=0`
`:. Deltaphi_(B)= phi_(1)= vecB_(1).vecA= B_(1) A cos theta` and here A =` 10 cm^(2) = 10^(-3) m^(2) ` and `theta = 0^(@)` i.e., cos `theta = 1` hence
Magnitude of magnetic field `B_(1) = (phi_(1))/(A cos theta)= (Deltaphi_(B))/(A cos theta)= (2.0)/(10^(-3)xx1)= 2.0xx10^(3)` T