Question

A solenoid is 1.5 m long and its inner diameter is 4.0 cm. It has 3 layers of windings of 1000 sums each and carries a current of 2.0 A. The magnetic flux for a cross-section of the solenoid Is nearly

• A. $4.1\times {10}^{-5}$ Wb
• B. $5.2\times {10}^{-5}$ Wb
• C. $6.31\times {10}^{-3}$ Wb
• D. $2.5\times {10}^{-7}$ Wb

Answer 1

The correct option is C $6.31\times {10}^{-3}$ Wb

Length of solenoid $l=1.5m$

Diameter of solenoid $d=4cm=0.04m$

Thus cross-sectional area of solenoid $A=\frac{\pi d^2}{4}$

$\therefore$ $A=\frac{3.14\times (0.04{)}^2}{4}=1.256\times {10}^{-3}{m}^2$

Total number of turns $N=3\times 1000=3000$ turns

Number of turns per unit length $n=\frac{N}{l}=\frac{3000}{1.5}=2000$

Magnetic field inside the solenoid $B={\mu }_{o}nI$

$\therefore$ $B=4\pi \times {10}^{-7}\times 2000\times 2=5.024\times {10}^{-3}Wb/m^2$

Magnetic flux through $1$ loop $\varphi =BA$

Thus magnetic flux through $1000$ loops ${\varphi }_t=1000BA$

$\therefore$ ${\varphi }_t=1000\times 2.054\times {10}^{-3}\times 1.256\times {10}^{-3}=6.31\times {10}^{-3}Wb$