A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 × 10⁻³ T and the angle of dip is 45°. The emf induced in the rod is ___________ mV.

Solution

Physics sir · 3 steps

Only the field cutting through the rod counts. Here that is the vertical component, so find B_V first.

1. Which component works. The horizontal component B_H points North, along the rod itself. A rod moving East through it gives a force on charges that points vertically, across the rod, so it drives no emf along the length. Only the vertical component B_V pushes charges along the rod.
2. Find B_V. From the angle of dip, tan δ = B_V/B_H. With δ = 45°, B_V = B_H tan 45° = 4 × 10⁻³ T.

   ε = B_V l v = (4 × 10⁻³)(0.2)(20) V

3. So. ε = 1.6 × 10⁻² V = 16 mV, so the answer is 16.

The common slip: 22.6. Using the total field B = B_H/cos 45° = 4√2 × 10⁻³ T instead of its vertical component gives about 22.6 mV, not 16.

The idea to keep. For motional emf, use only the part of B that is perpendicular to both the rod and its velocity. A horizontal rod moving horizontally cuts only the vertical component of Earth's field.

Answer from the NTA final answer key. Solution by PrepWiser.