Answer:
Explanation:
Given that,
Number of turn N = 40
Diameter of the coil d= 11cm = 0.11m
Then, radius = d/2 = 0.11/2 =0.055m
r = 0.055m
Then, the area is given as
A =πr²
A = π × 0.055²
A = 9.503 × 10^-3 m²
Magnetic Field B = 0.35T
Magnetic field reduce to zero in 0.1s, t = 0.1s
so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).
E.M.F is given as
ε = —N • dΦ/dt
Where magnetic flux is given as
Φ = BA
Then, ε = —N • dΦ/dt
ε = —N • dBA/dt
ε = —NBA/t
Then, its magnitude is
ε = NBA/t
Inserting the values of N, B, A and t
ε = 40×0.35×9.503×10^-3/0.1
ε = 1.33 V
Then, using the relationship between Electric field and electric potential
V = Ed
ε = E•d
E = ε/d
E = 1.33/0.11
E = 12.09 V/m
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Answer:
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