Question

A solenoid of length 2m and diameter 2mm carries 10 ampere
current. The magnetic field produced along its axis inside the
solenoid is 10mt. The length of the wire of the solenoid is ...​

Answer 1

Answer:

10m

Explanation:

Lenght of wire =

$\pi \: d(bl \div ui)$

put all the values in S. I

Answer 2

The number of turns present in the solenoid is 1592.

Given,

Length of the solenoid=2 m

Diameter of the solenoid=2 mm

Current through the solenoid=10 ampere

Magnetic field produced along its axis=10 milli-Tesla.

To find,

the number of turns of the wire of the solenoid(as the length of the solenoid is already given).

Solution:

• A solenoid has an arrangement in which an insulated wire is tightly wound closely in the form of a helix.
• It behave like a bar magnet.
• For an ideal solenoid(l>>R), the magnetic field inside it is uniform and outside it is zero.
• Magnetic field(B) along the axis of a solenoid is calculated as:
• $B=$µ0$ni$.
• where, µ0-permeabliity constant, n-number of turns per unit length and i-current.
• µ0=$4\pi X10^{-7}$.
• $n=\frac{N}{l}$.

As the length of the solenoid is much greater than the radius(diameter/2) of the solenoid, it is an ideal solenoid and the magnetic field along its axis wil be:

$Baxis=$µ0$ni$.

• $1mT=10^{-3} Tesla$.
• $10 mT=10X10^{-3} Tesla$.
• $10mT=10^{-2} Tesla$.

$Baxis=$µ0$ni$

$10^{-2} =4\pi X10^{-7}X nX10\\10^{-2}=4X3.14X10^{-6}Xn\\10^{-2}=12.56X10^{-6}Xn\\n=\frac{10^{-2}}{12.56X10^{-6}} \\n=\frac{10^{4}}{12.56} \\n=796.17\\n=796.2.$

The  n(number of turns per unit length) is 796.2.

$n=\frac{N}{l} \\N=nXl\\N=796.2X2\\N=1592.4.$

As the number of turns is a number, it will only be a whole number.

Thus, the number of turns in the solenoid is 1592.

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