Question A circular coil enclosing an area of 200 cm2 is made of 190 turns of copper wire. The wire making up the coil has resistance of 7.0 Ω, and the ends of the wire are connected to form a closed circuit. Initially, a 1.8-T uniform magnetic field points perpendicularly upward through the plane of the coil. The direction of the field then reverses so that the final magnetic field has a magnitude of 1.8 T and points downward through the coil. If the time required for the field to reverse directions is 0.50 s, what is the average current in the coil during that time? A

page-1Given:- Area of the coil =200 \mathrm{~cm}^{2}Number of turns =190Resistance of the wire =r=7 \Omegamagnetic field =|B|=1.8 \mathrm{~T}time =\Delta t=0.50 \mathrm{sec}Induced emf:- \epsilon=\left|-\frac{d \phi}{d t}\right| \quad \phi= flux (magnetic)change in magnetic field =\triangle B=B-(-B)=2 BEnf produced in one coil will bePage-2\begin{array}{l}\epsilon=\frac{d \phi}{d t}=A \frac{\Delta B}{\Delta t} \quad \quad(A=\text { Area) } \\t=200 \times 10^{-4} \times \frac{(2 \times 1.8)}{0.5} \\\epsilon=0.144 \mathrm{~V}\end{array}\Rightarrow Total emf produced in 190 coil will be\begin{array}{l}e m f=190 t=190 \times 0.144 \\e m f=27.36 \mathrm{~V}\end{array}The average current in the coil during this time will be\begin{array}{l}i=\frac{e m f}{\gamma}=\frac{27.36}{7} \\i=3.91 \mathrm{~A} \text { Answer. }\end{array}Answer: ...