Question

A magnetic field of strength 1.0 T is produced by a strong electromagnet in a cylindrical region of radius 4.0 cm, as shown in the figure. A wire, carrying a current of 2.0 A, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.
Figure

Answer 1

Explanation:

To find: The magnitude pf the force acting on the wire.

Given:

Magnetic field, (B)=1

Radius of the cylindrical region, r=4.0 cm

Electric current through the wire, I=2 A

The direction of magnetic field is perpendicular to the plane of the wire.

So, angle between wire and magnetic field, $\theta=90^{\circ}$

Magnetic Field,

$\begin{aligned}&\vec{F}=i \vec{l} \times \vec{B}\\&\therefore|\vec{F}|=i l l B \sin \theta\end{aligned}$

$|\vec{F}|=i l B \sin 90^{\circ}$

$Here, |=2 r$

$\begin{aligned}&\therefore|\vec{F}|=i 2 r B \sin 90^{\circ}\\&=2 \times 8 \times 10^{-2} \times 1.0 \times 1\\&=0.16 \mathrm{N}\end{aligned}$