Motional EMF of a Rod Moving Through a Tilted Magnetic Field: 04-BS-9, December 2016, Q6

Question 6: Motional EMF of a Rod Moving Through a Tilted Magnetic Field (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Rod length $L$ (N-S)	$1$ m
Velocity $v$	$5$ m/s, west
Field $B$	$10^{-5}$ T, $45^\circ$ above horizontal

Find. $V_{\text{north}}-V_{\text{south}}$.

Rod along N-S moves west; the local right-handed frame is East $=\hat x$, North $=\hat y$, Up $=\hat z$.

Approach. Motional EMF is $\int(\vec v\times\vec B)\cdot d\vec l$ along the rod. Set up a local right-handed East-North-Up frame; only the component of $\vec B$ perpendicular to BOTH the velocity and the rod (i.e. the vertical component) contributes an EMF along the N-S rod.

Isolate the field component that drives the EMF. With $\hat x=$East, $\hat y=$North, $\hat z=$Up (right-handed, $\hat x\times\hat y=\hat z$), $\vec v=-5\hat x$ (west) and the rod along $\hat y$. Only $B_z$ (the vertical component) produces a force along $\hat y$: $\vec v\times B_z\hat z=-5B_z(\hat x\times\hat z)=5B_z\hat y$. The field is $45^\circ$ above horizontal, so $$B_z=B\sin45^\circ=(10^{-5})(0.70711)=\boxed{7.071\times10^{-6}\ \text{T}}$$
Motional EMF along the rod. $$\varepsilon=(\vec v\times\vec B)\cdot\hat y\,L=vB_zL=(5)(7.071\times10^{-6})(1)$$ $$\varepsilon=\boxed{3.536\times10^{-5}\ \text{V}=35.36\ \mu\text{V}}$$
Polarity. The force on positive charge carriers, $q\vec v\times\vec B$, has its $\hat y$-component positive (Step 1), so positive charge is pushed toward the NORTH tip, making it the higher-potential end: $$\boxed{V_{\text{north}}-V_{\text{south}}=+3.536\times10^{-5}\ \text{V}\ \text{(north tip positive)}}$$

Quantity	Result
Vertical field component $B_z$	$7.071\times10^{-6}$ T
$V_{\text{north}}-V_{\text{south}}$	$+3.536\times10^{-5}$ V (north positive)

Check: the source does not state which horizontal direction the field's non-vertical component leans; this is immaterial here because the rod (N-S) and velocity (E-W) are both horizontal and orthogonal, so only $B$'s VERTICAL component enters the EMF (Step 1) — the horizontal component of $\vec B$, whatever its azimuth, contributes zero to this particular cross product.

Concept & Theory

Topic. Motional EMF, $\varepsilon=\int(\vec v\times\vec B)\cdot d\vec l$, for a rod moving through a uniform field with mutually orthogonal $v$, rod, and (relevant component of) $B$.

Key relations. $\vec F=q\vec v\times\vec B$ on the free charges; $\varepsilon=vBL$ when $v\perp B\perp L$ and all three are mutually orthogonal.

Why this works. A $45^\circ$ tilt is a common device in this exam: because $\sin45^\circ=\cos45^\circ$, the answer is identical whether the $45^\circ$ is measured from the horizontal or from the vertical, which quietly hides a convention ambiguity that would matter at any other angle — always confirm which reference axis a stated angle is measured from before trusting the shortcut.

Common pitfall. Using the FULL field magnitude $B$ instead of only its vertical component $B\sin45^\circ$ (the horizontal component of $\vec B$ contributes nothing: $\vec v\times\vec B_h$ is either zero, for $\vec B_h$ along E-W, or vertical, for $\vec B_h$ along N-S, so it has no component along the rod); or reporting the peak/magnitude without resolving the sign (polarity) the question explicitly asks for.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (motional EMF, the Lorentz force on free charges in a moving conductor).

Question 6 of 8: Motional EMF of a Rod Moving Through a Tilted Magnetic Field

Question 6: Motional EMF of a Rod Moving Through a Tilted Magnetic Field (20 marks)