Question

. 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

Answer 1

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