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# A conducting rod of length $l=2\;cm$ moves with velocity 12 m/s at an angle $30^{\circ}$ to a magnetic field produced between two square faced poles of side 2 cm. If flux leaving the face is $5 \mu Wb$ the induces emf is

$(a)\;30\;mv \\(b)\;15\;mv \\ (c)\;3.0\;mv \\ (d)\;1.5\;mv$

Induced $emf = Blv \sin 30^{\circ}$
$\phi = 5 \times 10^{-6} \;Wb$
$A= (2\;cm)^2= 4 \times 10^{-4}m^2$
$B=\large\frac{\phi}{\pi}$$=1.25 \times 10^{-2} Wb/m^2 \in = (1.25 \times 10^{-2}) \times (2 \times 10^{-2} ) \times 12 \times \large\frac{1}{2} \in = 3 \times \large\frac{1}{2}$$\times 10^{-3}$
$\quad= 1.5 \;mv$
Hence d is the correct answer.