To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,





At time
, Induced emf is,


Therefore the magnitude of the induced emf is 10.9V
Answer:
answer a: a large front gear with a small back gear
answer b: a small front gear with a large back gear
Explanation:
just simple gearing ratios
Answer:
Explanation:
For calculating resistance of a conductor , the formula is
R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.
For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂
Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.
R₁ / R₂ = I₁A₂ / I₂A₁
= 2l₂ x A₁ / 4 I₂A₁
= 1 / 2
2R₁ = R₂
Power = V² / R
Ratio of power = (V² / R₁) x (R₂ / V²)
= R₂ / R₁
= 2 .