A square shaped coil of area 70 cm² having 600 turns rotates in a magnetic field of 0.4 Wbm^-2, about an axis which is parallel to one of the sides of the coil and perpendicular to the direction of field. If the coil completes 500 revolutions in a minute, the instantaneous emf when the plane of the coil is inclined at 60° with the field, will be ______V. (Take $π = \frac{22}{7}$ )

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answer is 44.

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Detailed Solution

Given, $A = 70 \times 10^{- 4} m^{2}$
$\begin{array}{l} N = 600 \\ B = 0.4 Wb / m^{2} \\ ω = 50 \emptyset \times \frac{2 π}{6 \emptyset} rad / s = \frac{50 π}{3} rad / s \end{array}$
$θ = 30^{\circ} [∵$ Given angle 60° is between plane of coil and magnetic field]
$\begin{array}{l} \Rightarrow \sin θ = \sin 30^{\circ} = \frac{1}{2}; e = ? \\ e = BAN ω \sin θ = 0.4 \times 70 \times 10^{- 4} \times 600 \times \frac{50 π}{3} \times \frac{1}{2} \\ \Rightarrow e = 44 V \end{array}$