Question

A wire (mass = 55 g, length = 30 cm) is suspended horizontally by two vertical wires which conduct a current I = 5.0 A, as shown in the figure. The magnetic field in the region is into the paper and has a magnitude of 90 mT. What is the tension in either wire?

I I

Answer 1

Solution:

The mass of the wire, $m=55\; \mathrm{gm}=55\times 10^{-3}\; \mathrm{kg}$

Length of the wire, $L=30\; \mathrm{cm}=30\times 10^{-2}\; \mathrm{m}$

The current passing through the conductor,

i=5A

The magnetic field in the region,

B = 90 mT = 90 x 10-3 T

Let us consider the tension in each wire is

Now, we will check the forces acting on the conductors.

The weight $(F_{w}=mg)$ and magnetic force $(F_{m}=BiL)$ will act in a downward direction.

Total tension in the wires is $2T\; \; \; \; \; \; \; \; \; \; \; (\because \mathrm{tension\; in\; each\; wire\; is\; T\; and\: we\: have\: two\: wires})$

These two forces will be balanced

Therefore, we have

$2T=F_{w}+F_{m}$

$2T=mg+BiL$

$T=\frac{mg+BiL}{2}$

Now, substitute all the known values

$T=\frac{(55\times 10^{-3}\times 9.8)+(90\times 10^{-3}\times 5\times 30\times 10^{-2})}{2}$

$T=\frac{0.539+0.135}{2}$

${\color{Blue} T=0.337\; \mathrm{N}}$

This is the tension in each wire.