Question

Problem 3 A wire of length 16cm is suspended by flexible leads above a long straight wire. Equal but opposite currents are established in the wires such that the 16cm wire floats 1.5mm above the long wire with no tension in the suspension leads. If the mass of the 16cm wire is 14g = 0.014kg, what is the current? Select One of the Following: (a) 16A (b) 80A (c) 120A (d) 3.2A (e) 0.13A

Answer 1

Answer:

Option B

Solution:

As per the question:

Length of the suspended wire, l = 16 cm = 0.16 m

Distance between the long wire and the suspended wire, R = 1.5 mm = 0.0015 m

Mass of the 16 cm wire, m = 0.014 kg

Now,

We know that the magnetic field due to the two parallel current carrying straight long wires with current 'I' and the separation 'R' between the wires is given by:

$B = \frac{\mu_{o}I}{2\pi R}$                  (1)

If the suspended wire floats, force due to its weight is balanced by the magnetic force due to current I:

mg = IBl

Also, from eqn (1):

mg = $I\frac{\mu_{o}I}{2\pi R}l$

mg = $\frac{\mu_{o}I^{2}l}{2\pi R}$

$0.014\times 9.8 = I\frac{4\pi \times 10^{- 7}I}{2\pi \times 0.0015}\times 0.16$

$I^{2} = 6430.436$

I = 80.19 A

I = 80 A (approx)

Answer 2

Answer:

I = 80 A

Explanation:

It is given that,

Mass of the wire, m = 14 g = 0.014 kg

Length of the wire, l = 16 cm = 0.16 m

Distance from wire, $r = 1.5\ mm = 1.5\times 10^{-3}\ m$

The force of gravity acting on the wire is balanced by the magnetic force. Mathematically, it is given by :

$mg=ILB$

The magnetic field at the axis of wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{mg}{LB}$

$I=\dfrac{2\pi r mg}{L \mu_o I}$

$I^2=\dfrac{2\pi \times 1.5\times 10^{-3} \times 0.014\times 9.8}{0.16\times 4\pi \times 10^{-7}}$

$I^2=6431.25$

I = 80 A

So, the current flowing in the wire is 80 A. Hence, this is the required solution.