202212020955 Solution to 1978-HL-PHY-II-4

A rectangular loop of length 0.2\,\mathrm{m} and width 0.1\,\mathrm{m}, carrying a steady current I of 2\,\mathrm{A} is hinged along the y-axis, and is situated in a uniform magnetic field \mathbf{B} of 0.5\,\mathrm{T} parallel to the x-axis.

If the plane of the loop makes an angle of 60^\circ with the xy plane,

(a) calculate the force exerted by the magnetic field on each side of the loop, and
(b) calculate the torque required to hold the loop in this position.


Roughwork.

Force on a current-carrying conductor in a magnetic field is in magnitude

\boxed{F=BIl\sin\theta}

its direction to be determined by Fleming’s left hand rule.

(a) Have in mind a picture as viewing cross-sectionally:

and as down the top:

where

\begin{aligned} F_1 = F_3 & = (0.5)(2)(0.2)\sin 90^\circ \\ F_2 = F_4 & = (0.5)(2)(0.1)\sin 60^\circ \\ \end{aligned}

(b) This part is not to be attempted.