Question

A wire carries a 4.0-A current along the +x-axis through a magnetic field B= (5.0 i+7.0 j) T. If the wire experiences a force of 30 N as a result, how long is the wire?

Answer 1

Hello!

The equation for the magnetic force experienced by a wire is:

`F_B = B * iL`

F = Magnetic force on wire (30 N)

B = Magnetic field strength (5.0i + 7.0j)

i = Current in wire (4.0 A)
L = Length of wire (? m)

This is a CROSS-PRODUCT. We can rewrite this as:

`F_B = BiLsin\theta`

Where 'θ' is the angle between the magnetic field and the current carrying wire.

Therefore, we would NOT count any field that is parallel to the wire. The field does have a component (5.0i) that is along the x-axis, so we would NOT count it because the wire is along this same axis.

This leaves the 'j' component, which is PERPENDICULAR to the wire. θ would then be 90°, and sin(90) = 1.

`F_B = B\hat{j} iL`

Solving for the length:

`L = \frac{F_B}{B\hat{j} i}\\\\L = (30)/(7(4)) = \boxed{1.07 m}`