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. 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

Respuesta :

akindeleot

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 = [tex]\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right][/tex] = (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

onyebuchinnaji

The resulting magnetic force on the wire is 1,200 N.

Resulting magnetic force

The resulting magnetic force is calculated as follows;

F = BIL

F = I (B x L)

F = I [(30i, -40j) x (2i, 0)]

[tex]\left[\begin{array}{cc}i&j\\30&-40\\2&0\end{array}\right] \\\\B\times L = 0 - (-80) = 80[/tex]

F = I(B x L)

Where;

I is the current in the wire

F = 15 x 80

F = 1200 N

Thus, the resulting magnetic force on the wire is 1,200 N.

Learn more about magnetic force here: https://brainly.com/question/13277365

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