A solenoid 60 cm long and of radius 4 cm has 3 layer of windings 300 turns each. A 2.3 cm long wire of mass 2.5g lies inside the solenoide near its centre normal to its axis, both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire?

Options

A. `105.27`
B. `210`
C. `108.27`
D. `110`

Answer

C

Explanation

Let I be the current in the windings of the solenoid that can
support the weight of the wire. Then, magnetic field inside the
solenoid along its axis is given by `B=mu_0nI`.
Given that : `mu_0=4pixx10^-7T mA^_1`
n=number of turns per unit length =`(3xx300)/(60 cm)=(3xx300)/(60xx10^-2m)`
`=1500 turns//metre`
Hence, `B=4pixx10^07xx1500xxI=6pixx10^-4I`
As the wire is placed normal to the axis of the solenoid and
magnetic field inside the solenoid is along the axis, hence field
due to the magnetic field is given by: `F=BI'l`
Here, `I'=6.0A, l=2cm=0.2xx10^-2m, and B=6pixx10^-4I`
Hence, `F=6pixx10^-4xx6.0xx2.0xx10^-2N=72pixx10^-6I`
Current I will support the weight of the wire if the force F equal
the weight of the wire, i.e. `F=mg`
or `72pixx10^-6I=2.5xx10^-3xx9.8`
`implies I=(2.5xx10^-3xx9.8)/(72xx3.14xx10^-6)=108.37A`