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2 years ago

5

. A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and

given by B = (30 − 40 ) mT. What is the resulting magnetic force on the wire

1 answer:

Lady bird [3.3K]2 years ago

6 0

Answer:

The resulting magnetic force on the wire is -1.2kN

Explanation:

The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is

F = I (L*B)

Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)

L x B = $\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right]$ = (0, 0, -80)

we can now solve

F = I (L x B) = I (-80)

F = -1200 kmN

F = -1200 kN * 10⁻³

F = -1.2kN

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Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde

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Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

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2 years ago

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Answer:

$\frac{1}{10}$ M

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$I_{1}$ is the moment of interia of earth before impact
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let $W_{1}=W$

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where M is mass of earth

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$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

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$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

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⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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