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kozerog [31]
3 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
Physics
1 answer:
Lady bird [3.3K]3 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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Svetllana [295]

Answer:

a) \Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}, b) \Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}, c) \Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}

Explanation:

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b) The flux through the right face is:

\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

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