Electromagnetism — Question 256

ANSWER

(a) Left end N, right end S; internal field points left. (b) $2.01\times10^{-3}\,\mathrm T$.

Worked solution

(a) Counterclockwise current as seen from the left makes the field point toward that viewer. That end is N; the internal field runs from right S to left N.

(b) $n=N/L=900/0.450=2000\,\mathrm{m^{-1}}$. Thus $B=\mu_0nI=4\pi\times10^{-7}(2000)(0.800)=2.01\times10^{-3}\,\mathrm T$.

Equivalent physically justified methods, alternative valid sketches and consistent units accepted.

Accept sensible significant figures and consistent rounding from stated constants.

Part a: Identify left N and right S.

[1 mark]

Part a: Draw internal field arrows toward the left.

[1 mark]

Part b: Use the number of turns per metre.

[1 mark]

Part b: Calculate the field with units.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.