What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s

Induced peak emf = NABω

= 500 x (π x 0.25²) x 0.425 x 376.738

= 15721.16 V

= 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0

2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

$\bar{K}=\frac{3k_{B}T}{2}$

Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

After the absorption,

$(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}$ (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

$\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

$\frac{v_{2}}{v_{1}}=\sqrt{4}$

Therefore the ratio will be $\frac{v_{2}}{v_{1}}=2$

I hope it helps you!

4 0

3 years ago

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