Current through an inductor

Assume we have an inductor with the following values:
$$B=B_0 \cdot sin(2\cdot \pi \cdot f \cdot t)$$ $$N=10$$ $$r=5cm$$ $$B_0 = 2mT$$ $$f=50Hz$$ $$R=5 \Omega$$ How much current runs through the inductor as function of time?
Well I don't know the inductance, so I tried this:
$$U=-\frac{d \phi}{dt}$$ $$\phi = B \cdot A = 2mT \cdot sin(\pi \cdot 100 \cdot t) \cdot \pi \cdot (2\cdot10^{-2})$$ $$U = -\frac{d}{dt} 2mT \cdot sin(\pi \cdot 100 \cdot t) \cdot \pi \cdot (2\cdot10^{-2})^2$$ $$U = -\frac{d}{dt} 8 \cdot \pi \cdot 10^{-7}T \cdot sin(\pi \cdot 100 \cdot t)$$ $$U = -8 \cdot \pi \cdot 10^{-7}T \cdot \frac{d}{dt} sin(\pi \cdot 100 \cdot t)$$ $$U = -8 \cdot \pi \cdot 10^{-7}T \cdot 100\cdot \pi \cdot cos(\pi\cdot 100\cdot t)$$ $$U = -800 \cdot \pi^2 \cdot 10^{-7}T \cdot cos(\pi\cdot 100\cdot t)$$ $$I=\frac{U}{R}= - \frac{800 \cdot \pi^2 \cdot 10^{-7}T \cdot cos(\pi\cdot 100\cdot t)}{5\Omega} = -1.579\cdot 10^{-4}\cdot cos(\pi\cdot 100\cdot t)A$$

I never saw something like that anywhere else so I'm a bit unconfident if this approach was really correct. Is this correct? Is there maybe another (easier) approach?

• $\Phi _{\mathbf {B} }=LI$ You must compute L en.wikipedia.org/wiki/Inductor#Inductance_formulas using B – Tony Stewart Sunnyskyguy EE75 Jan 27 at 20:06