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.

Question

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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e23e3e617977ed44b08e0417e42ce29f-.png" /> B H =4 × 10-3 T θ arrow 45° B V = B H ϵ=( V × B ) ⋅ vecℓ =((4 × 10-3)(20)) (20/100) =16 × 10-3 V =16 mV

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

BH​=4×10−3T θ→45∘ BV​=BH​ ϵ=(V×B)⋅ℓ =((4×10−3)(20))10020​ =16×10−3V=16mV

$B_{H} = 4 \times 1 0^{- 3} T$
$θ \to 4 5^{\circ}$
$B_{V} = B_{H}$
$ϵ = (V \times B) \cdot ℓ$
$= ((4 \times 1 0^{- 3}) (20)) \frac{20}{100}$
$= 16 \times 1 0^{- 3} V = 16 mV$