= Solution
Choose a positive unit normal $\hat{\mathbf n}$ to the loop and choose positive circulation by the <right-hand rule>. Write the applied <magnetic field> as $\mathbf B(t)=B_0\cos(\omega t)\hat{\mathbf b}$, where $\hat{\mathbf b}\cdot\hat{\mathbf n}=\cos\theta$. Because the field is uniform, the external <magnetic flux> is
$$
\Phi_B(t)=\int\mathbf B\cdot d\mathbf S
=\pi r^2B_0\cos\theta\cos(\omega t).
$$
<Faraday's law of induction> gives the induced <electromotive force>, and <Ohm's law> gives the current through the loop's <electrical resistance>:
$$
\mathcal E=-\frac{d\Phi_B}{dt}=\pi r^2B_0\omega\cos\theta\sin(\omega t),
\qquad
\boxed{I(t)=\frac{\pi r^2B_0\omega\cos\theta}{R}\sin(\omega t).}
$$
This is the <induced current in a tilted resistive loop> with the self-generated magnetic-flux contribution neglected as requested. The factor has units $\mathrm{Wb\,s^{-1}\,\Omega^{-1}}=\mathrm A$. A field parallel to the loop plane has $\cos\theta=0$ and induces no current in this uniform-field model.
For $\cos\theta>0$, the initial normal flux is positive. At $\omega t=\pi/2$, the external <magnetic field> is instantaneously zero but its normal component is decreasing, so $d\Phi_B/dt<0$ and $I>0$. The current is counterclockwise when viewed from the positive-normal side. Its own <magnetic field> points along $+\hat{\mathbf n}$ through the loop, preserving the initial normal flux in accordance with <Lenz's law>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019-ia-physics-q5-induction-directions.png]
{title=Initial oblique field and the induced current a quarter-cycle later, for a positive initial normal flux}
For $\cos\theta<0$, both circulation and the induced normal-field direction reverse. The current follows the rate of change of <magnetic flux>; the quarter-cycle current is nonzero even though the external field itself vanishes then.
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