Question

The length of wingspan of an aeroplane is \( L \) meters and it is flying with a velocity of \( v \, \text{m/s} \) from north towards south. If the horizontal component of Earth's magnetic field is \( H \, \text{weber/m}^2 \) and induced e.m.f. produced between the ends of the wingspan is \( e \, \text{volt} \), then obtain the expression for the angle of dip at that place.

Hint:

The induced e.m.f. depends on the velocity of the object, the length of the conductor, and the magnetic field strength. The angle of dip can be determined by the induced e.m.f.

Answer 1

Step 1: Formula for the Induced E.M.F.
The induced e.m.f. \( e \) due to the motion of the aeroplane in Earth's magnetic field is given by: \[ e = B L v \sin \theta \] where: - \( B \) is the magnetic field strength, - \( L \) is the wingspan of the aeroplane, - \( v \) is the velocity of the aeroplane, and - \( \theta \) is the angle of dip (the angle between the magnetic field and the horizontal).
Step 2: Horizontal Component of the Magnetic Field.
The horizontal component of Earth's magnetic field is \( H \), so we have: \[ B = H \]
Step 3: Substituting into the Formula.
The induced e.m.f. becomes: \[ e = H L v \sin \theta \]
Step 4: Solving for \( \theta \).
Rearranging the above equation to solve for \( \theta \), we get: \[ \sin \theta = \frac{e}{H L v} \] Thus, the expression for the angle of dip \( \theta \) is: \[ \theta = \sin^{-1} \left( \frac{e}{H L v} \right) \]
Step 5: Conclusion.
Thus, the angle of dip at the place is given by: \[ \boxed{\theta = \sin^{-1} \left( \frac{e}{H L v} \right)} \]