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3 years ago

6

You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

ngth is 2.47 cm. What is the inductance, in microhenries, of your solenoid

1 answer:

lina2011 [118]3 years ago

3 0

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

$L = \frac{\mu N^2 A}{l}$

Here,

$\mu$ = Permeability at free space

N = Number of loops

A = Cross-sectional Area

l = Length

Replacing with our values we have,

$L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}$

$L = 0.00002442H$

$L = 24.42\mu H$

Therefore the Inductance is $24.42\mu H$

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(a) The two balls collide $2\; \rm s$ after launch.

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Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

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On the other hand, the initial velocity of the ball thrown upwards should be positive: $\left(12\; \rm m \cdot s^{-1}\right)$ .

Let $v_0$ and $h_0$ denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time $t$ :

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Initial height: $h_0 = 0\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + 12\, t$ (where $h$ is in meters and $t$ is in seconds.)

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$-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t$ .

$t = 2$ .

Therefore, the collision takes place $2\, \rm s$ after launch.

Substitute $t = 2$ into either of the two original expressions to find the height of collision:

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