9.1.1. A force of $15 \mathrm{~N}$ acts on a linear conductor of length $10\mathrm{~cm}$, located perpendicular to a magnetic field, if the current in the conductor is $1.5\mathrm{~A}$. Find the induction of the magnetic field.
Solution
We find the Ampere force as $$\vec{F} = (\vec{Il})\times \vec{B}$$ In scalar form, this expression can be rewritten as $$F = B\cdot Il \cdot \sin\alpha\tag{1}$$ since the field is perpendicular to the current element, then $$\alpha =90^{\circ};\quad \sin\alpha =1$$ From here, we rewrite the expression $(1)$ as $$F = B \cdot Il \Rightarrow \boxed{B = \frac{F}{Il} = 100\text{ T}}$$