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Ede4ka [16]
3 years ago
15

A wire carrying a 32.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's

field, and there is a 2.15 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?
Physics
1 answer:
marin [14]3 years ago
7 0

Answer:

2.24 T

Explanation:

From Electromagnetic Field,

F = BILsin∅................ Equation 1

Where F = Force on the wire, B = Field strength, I = current flowing in the conductor, L = length of the conductor, ∅ = The angle the conductor makes with the magnetic field.

Making B the subject of the equation,

B = F/ILsin∅..................... Equation 2

Given: F = 2.15 N, I = 32 A, L = 3.00 cm = 0.03 m, ∅ = 90° ( the wire is perpendicular to the magnetic field)

Substitute into equation 2

B = 2.15/(32×0.03×sin90°)

B = 2.15/0.96

B = 2.24 T.

Hence the Field strength = 2.24 T

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Answer:

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A = (8400 Bq) (½)^(18.0 min / 4.50 min)

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Two small nonconducting spheres have a total charge of 94.0 μC . Part A
Leto [7]

Answer:

Part A;

The charges are;

Q₁ = 1.32687 μC and Q₂ = 92.67313 μC

Part B

Q₁ = 95.29042 μC, Q₂ = -1.29042 μC

Explanation:

Part A

The total charge on the two nonconducting spheres = 94.0 μC

The force exerted by each on the other when placed 31 cm apart = 11.5 N

Let Q₁ = x represent the charge on one of the spheres and let Q₂ represent the charge on the other sphere

The force, 'F', exerted by a charge is given as follows;

F = k \times \dfrac{\left | Q_1  \right | \cdot \left | Q_2  \right |  }{d^2}

Where;

d = The distance between the spheres = 31 cm = 0.31 m

k = 8.9875 × 10⁹ N·m²/C²

Where F = 11.5 N

Q₁ + Q₂ = 94.0 μC

∴ Q₂ = 94.0 - Q₁ = 94.0 - x

We get;

F =11.5 =  8.9875 \times 10^9 \times \dfrac{x\times (94.0-x) \times 10^{-12}}{0.31^2}

Therefore;

94·x - x² - 122.965 = 0

x² - 94·x + 122.965

x = (94 ± √((-94)² - 4×1×122.965))/(2 × 1)

Solving gives;

x ≈ 1.32687 × 10⁻⁶ C or x = 92.67313 × 10⁻⁶ C

Therefore, the charges are;

Q₁ = 1.32687 × 10⁻⁶ C and Q₂ = 92.67313 × 10⁻⁶ C

Q₁ = 1.32687 μC and Q₂ = 92.67313 μC

Part B

For attractive force, we have;

Q₁ + Q₂ = 94 × 10⁻⁶...(1)

11.5 =  8.9875 \times 10^9 \times \dfrac{-x\times (94.0-x)}{0.31^2} =  8.9875 \times 10^9 \times \dfrac{-Q_1\times Q_2}{0.31^2}

-Q₁ × Q₂ = 11.5 × 0.31²/(8.9875 × 10⁹) = 1.2296523 × 10⁻¹⁰...(2)

∴ Q₂ = -1.2296523 × 10⁻¹⁰/(Q₁)

Q₁ + Q₂ = Q₁  - 1.2296523 × 10⁻¹⁰/(Q₁) = 94 × 10⁻⁶

Q₁² - 94 × 10⁻⁶·Q₁ - 1.2296523 × 10⁻¹⁰ = 0

∴ Q₁ = (94 × 10⁻⁶ ± √((-94 × 10⁻⁶)² - 4 × 1 × 1.2296523 × 10⁻¹⁰))/(2×1)

Q₁ = 9.529042 × 10⁵ C or -1.29042 × 10⁻⁶ C

Therefore, Q₁ = 9.529042 × 10⁵ C and Q₂ = -1.29042 × 10⁻⁶ C

Q₁ = 95.29042 μC and Q₂ = -1.29042 μC

7 0
2 years ago
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