A circular coil enclosing an area
of 200 cm^{2} 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

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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: ...