Given emf E = 2 V, internal resistance r = 0.5 ohm, external resistance R = 4.5 ohm. Current I = E/(R + r) = 2/(4.5 + 0.5) = 2/5 = 0.4 A. Terminal potential difference V = I R = 0.4 x 4.5 = 1.8 V (equivalently V = E - I r = 2 - 0.4 x 0.5 = 1.8 V).

I = E/(R + r); V = E - I r

The force on a current-carrying conductor is F = B I L sin theta. Here B = 0.2 T, I = 4 A, L = 0.5 m and theta = 90 deg (so sin theta = 1). F = 0.2 x 4 x 0.5 x 1 = 0.4 N. The direction of the force is given by Fleming's left-hand rule.

F = B I L sin theta

Consider a circular loop of radius R carrying current I. By the Biot-Savart law, each element I dl is perpendicular to the line joining it to the centre (theta = 90 deg), so dB = (mu0/4 pi)(I dl)/R^2. All elements produce fields in the same direction (perpendicular to the plane of the loop) at the centre, so they simply add. Integrating dl around the loop gives the circumference 2 pi R: B = (mu0/4 pi)(I/R^2)(2 pi R) = mu0 I/(2R). For N turns, B = mu0 N I/(2R).

B = mu0 I/(2R)