Question

An aeroplane with its wings, spread $10m$, is flying at a speed of $180km/h$ in a horizontal direction. The total intensity of the magnetic field at that part is $2.5\times {10}^{–4}Wb/m^2$ and the angle of dip is ${60}^{°}$. The emf induced between the tips of the plane wings will be:

• A. $88.37mV$
• B. $62.50mV$
• C. 54.125 mV
• D. $108.25mV$

Answer 1

The correct option is D

$108.25mV$

Explanation for the correct option:

Step 1: Given data:

Length of its wings, $l=10m$

Speed, $v=180km/h=180x\frac{5}{18}m/s=50m/s$

Intensity of magnetic field, $B=2.5\times {10}^{–4}Wb/m^2$

Angle of dip, $\theta ={60}^{°}$

Step 2: Finding the induced emf:

We know that, $\sin \left(\theta \right)=\frac{B_{vertical}}{B}$

$B_{vertical}=\sin \left(60\right)x2.5x{10}^{-4}$

Using the formula for Induced emf in a conductor placed in magnetic field,

$$\begin{gathered} E=B_{vertical}lv \\ =\frac{\sqrt{3}}{2}x2.5x{10}^{-4}x10x50 \\ =10.825\times {10}^{-2} \\ =108.25mV \end{gathered}$$

Hence, option D is correct.