Question

A part of a wire carrying 2.0 A current and bent at 90° at two points is placed in a region of uniform magnetic field `vec B = -(0.50 T)vec k`, as shown in the figure. Calculate the magnitude of the net force acting on the wire.

Answer 1

From the figure:

Vertical segment length = 20 cm = 0.20 m

Horizontal displacement = 50 cm = 0.50 m

Magnetic field `(vec B) = -0.50 hat k T`

The wire effectively forms an L-shape.

Net displacement vector `(vec L_"net") = 0.50 hat i - 0.20 hat j`

Net force `(vec F) = I(vec L_"net" xx vec B)`

= `2[(0.50 hat i - 0.20 hat j) xx (-0.50 hat k)]`

Cross product:

`hat i xx hat k = -hat j`

`hat j xx hat k = hat i`

`(0.50 hat i) xx (-0.50 hat k) = 0.25 hat j`

`(-0.20 hat j) xx (-0.50 hat k) = 0.10 hat i`

So,

`vec F = 2(0.10 hat i + 0.25 hat j)`

= `0.20 hat i + 0.50 hat j`

Magnitude (F) = `sqrt(0.20^2 + 0.50^2)`

= `sqrt(0.04 + 0.25)`

= `sqrt 0.29`

F ≈ 0.54 N