WAEC Physics 2011 Paper

Concept:The emf induced in a conductor moving perpendicular to a magnetic field is given by $\mathcal{E} = B L v$.Explanation:Convert the length of the wire into metres: $L = 30\text{ cm} = 0.30\text{ m}$.The speed of the wire is $v = 2\text{ ms}^{-1}$.The magnetic flux density is $B = 0.4\text{ T}$.Since the wire moves at right angles to the magnetic field, the angle $\theta = 90^\circ$ and $\sin 90^\circ = 1$.Using the formula for motional emf:$$\mathcal{E} = B L v \sin \theta = 0.4 \times 0.30 \times 2 \times 1 = 0.24\text{ V}$$Thus, the induced emf in the wire is $0.24\text{ V}$.Answer:The correct option is B: $0.24\text{ V}$.