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Bad White [126]
2 years ago
5

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.50 x 10-4 T) at a distance of 15

cm from the wire, what is the maximum current the wire can carry? Express your answer using two significant figures.
Physics
1 answer:
ser-zykov [4K]2 years ago
3 0

Answer:

I = 37.5\ A

Explanation:

Given,

Magnetic field,B = 0.5 x 10⁻⁴ T

distance,r= 15 cm = 0.15 m

Current = ?

Using Ampere's law of magnetic field

B = \dfrac{\mu_0I}{2\pi r}

I= \dfrac{B (2\pi r)}{\mu_0}

I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}

I = 37.5\ A

Current in the wire is equal to I = 37.5\ A

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A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
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Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

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       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
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       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

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       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

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  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

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