On a straight conductor 80 in length placed in a uniform magnetic field from the side of the magnetic field

On a straight conductor 80 in length placed in a uniform magnetic field from the side of the magnetic field, a force equal to 0.2 H? determine the current strength and the direction of the current in the conductor if the magnetic field induction is 0.04.

According to Ampere’s law:
F = I * B * L * sin a, where F = 0.2 N (from the conditions of the problem), I is the current strength, B is the value of the magnetic field induction (B = 0.04 T), L is the length of the conductor (L = 80 cm = 0.8 m), sin a = 1 (since the conductor is straightforward, then a = 90 and sin a = sin 90 = 1).
I = F / (B * L * sin a) = 0.2 / (0.04 * 0.8 * 1) = 0.2 / 0.032 = 6.25 A.
Check: A = H / (Tl * m) = (kg * m / s ^ 2) / ((kg * s ^ -2 * A ^ -1) * m) = 1 / (A ^ -1) = A …
Answer: the current strength is 6.25 A; the direction of the current is perpendicular to the plane passing through the axis of the conductor.