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Question: A solenoid is \(1.5\,m\) long and its inner diameter is \(4.0\,cm\). It has three layers of windings...

A solenoid is 1.5 m1.5\,m long and its inner diameter is 4.0 cm4.0\,cm. It has three layers of windings of 10001000 turns each and carries a current of 2.0 amperes2.0\,amperes. The magnetic flux for a cross-section of the solenoid is nearly.

A) 4.1×10−5 weber4.1 \times {10^{ - 5}}\,weber

B) 5.2×10−5 weber5.2 \times {10^{ - 5}}\,weber

C) 6.31×10−3 weber6.31 \times {10^{ - 3}}\,weber

D) 2.5×10−7 weber2.5 \times {10^{ - 7}}\,weber

Explanation

Solution

Here we have to use the formula of cross-sectional area of solenoid and then use the formula of magnetic flux.

Magnetic flux is defined as the number of magnetic field lines flowing through the closed surface. It is used to calculate the overall magnetic field that passes through a given surface.

Magnetic flux is the cause of the field surrounding the magnetic material. It consists of photons, but unlike the light we obtain from the Sun, it is at a much lower level.

Complete step by step solution:
Magnetic field is a vector field that defines the magnetic effect on the passage of electrical charges, electrical currents and magnetised objects. A charge that travels in a magnetic field encounters a force perpendicular to its own momentum and to the magnetic field.

The Biot-savart law can be used to calculate the magnetic field strength of the current section.
Given,
Length of the solenoid, l=1.5 ml = 1.5\,m

Diameter of the solenoid d=4 cm=0.04 md = 4\,cm = 0.04\,m

Cross-sectional area of the solenoid is given by:
A=πd24 =3.14×(0.04)24 =1.256×10−3 m2  A = \dfrac{{\pi {d^2}}} {4} \\\ = \dfrac{{3.14 \times {{\left( {0.04} \right)}^2}}} {4} \\\ = 1.256 \times {10^{ - 3}}\,{m^2} \\\

Total number of turns,
N=3×1000=3000N = 3 \times 1000 = 3000

Number of turns per unit length,
n=Nl =30001.5 =2000  n = \dfrac{N} {l} \\\ = \dfrac{{3000}} {{1.5}} \\\ = 2000 \\\

Biot-Savart law is given by:
B=μ∘nI =4π×10−7×2000×2 =5.024×10−3 Wbm−2  B = {\mu _ \circ }nI \\\ = 4\pi \times {10^{ - 7}} \times 2000 \times 2 \\\ = 5.024 \times {10^{ - 3}}\,Wb{m^{ - 2}} \\\

Magnetic flux is given by:
ϕ=BA\phi = BA

Since, the magnetic flux passes through 10001000 loops, then
ϕ=1000×2.054×10−3×1.256×10−3 =6.31×10−3 Wb  \phi = 1000 \times 2.054 \times {10^{ - 3}} \times 1.256 \times {10^{ - 3}} \\\ = 6.31 \times {10^{ - 3}}\,Wb \\\

Hence, option C is correct.

Thus, the magnetic flux for a cross-section of the solenoid is nearly 6.31×10−3 Wb6.31 \times {10^{ - 3}}\,Wb .

Note:
Here we have to see whether the number of turns is given or not. Also, we have to observe whether length of material is given or not. If given then, we have to divide the number of turns by the length to get the number of turns per unit length of the material.