$\begin{array}{1 1} 0.1994\;m \\ 0.1894\;m \\ 0.5994 \;m \\ 0.0359\;m\end{array}$

- Electromagnetic induction $\varepsilon = N \large\frac {d\Phi}{dt}$

Electromagnetic induction $\varepsilon = N \large\frac {d\Phi}{dt}$$ = N \large\frac {\Delta B A \cos \theta}{\Delta t }$ $ = NA\cos \theta \large\frac{B_2 - B_1}{\Delta T}$

Given $N = 10, \theta = 13 ^{\circ}, r = 12 \times 10^{-2} m, B_1 = 2 T, B_2 = 5T, \Delta t = 12 s$

$\varepsilon = 10 \times \pi \times (12 \times 10^{-2})^2 \times \cos 13 \times \large\frac{3-5}{12}$

$\varepsilon \approx -0.0734 V$

Now, If the angle is changed to $67 ^{\circ}$, we need to caclulate the radius, assuming the emf stays the same.

$\Rightarrow -0.0734 = 10 \times \pi \times (r)^2 \times \cos 67 \times \large\frac{3-5}{12}$

$\Rightarrow -0.0734 = -2.0448 r^2$

$\Rightarrow r = \sqrt 0.03589$

$\Rightarrow r \approx 0.1894 m$

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