Answer
$B =\frac{ QR}{N(2A)} $
Work Step by Step
The average induced emf is given by Faraday’s law, equation 21–2b. The coil is rotated through a half-turn, so the initial flux changes sign. Therefore, the magnitude of the change in flux is twice the initial flux.
$$emf_{average}= N\frac{\Delta \Phi}{\Delta t}=\frac{N(2AB)}{\Delta t}$$
The average current is the average induced emf divided by resistance R. Finally, the charge Q that flows is the current multiplied by the elapsed time.
$$Q=I\Delta t=\frac{ emf_{average}}{R}\Delta t=\frac{\frac{N(2AB)}{\Delta t}}{R}\Delta t$$
$$Q =\frac{N(2AB)}{ R} $$
$$B =\frac{ QR}{N(2A)} $$
This was to be shown.
