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Nezavi [6.7K]
2 years ago
14

A Circular loop in the plane of a paper lies in a 0.65 TT magnetic field pointing into the paper. The loop's diameter changes fr

om 17.5 cmcm to 6.6 cmcm in 0.48 ss .
a) Determine the direction of the induced current and justify your answer.b) Determine the magnitude of the average induced emf.c) If the coil resistance is 2.5 Ω, what is the average induced current?
Physics
1 answer:
Kobotan [32]2 years ago
6 0

Answer:

Explanation:

magnetic filed, B = 0.65 T

initial diameter, d = 17.5 cm

final diameter, d' = 6.6 cm

time, t = 0.48 s

(a) According to Lenz's law, the direction of induced current is clockwise.

(b) Let e is the induced emf.

initial area, A = π r² = 3.14 x 0.0875 x 0.0875 = 0.024 m²

final area, A' = π r'² = 3.14 x 0.033 x 0.033 = 0.00342 m²

change in area, ΔA = A - A' = 0.024 - 0.00342 = 0.02058 m²

The magnitude of induced emf is given by

e=\frac{d\phi }{dt}

e=B\frac{\Delta A }{dt}

e = 0.65 x 0.02058 / 0.48

e = 0.028 V

(c) R = 2.5 ohm

i = e / R

i = 0.028 / 2.5

i = 0.011 A

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Initially, basket ball with the 13 rules was played with 2 peach baskets setup as goals, which now a days are the baskets on a high poles although modified but with same basic idea. In the very first game played in Springfield, the set of players were able to score a single point only, in the whole game.

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Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

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ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

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