Question

A solenoid 1.5 m long and 4.0 cm in diameter possesses 10 turns/ cm. A current of 5 A is flowing through it. Then the magnetic induction (i) inside and (ii) at one end on the axis of solenoid are respectively

• A. $2\pi \times {10}^{-3}T,\pi \times {10}^{-3}T$
• B. $3\pi \times {10}^{3}T,1.5\pi \times {10}^{3}T$
• C. $0.5\pi \times {10}^3{10}^{3}T,4\pi \times {10}^{3}T$
• D. $4\pi \times {10}^{3}T,2\pi \times {10}^3$

Answer 1

The correct option is B $3\pi \times {10}^{3}T,1.5\pi \times {10}^{3}T$

Given that,

Length of solenoid = $1.5$ m

Numbers of turns $n=10$ turns/cm

$n=10\times 150=1500$ turns

Current I = $5$ A

We know that,

Magnetic induction at the axis inside the solenoid

$B={\mu }_{0}Ni$

$B=4\pi \times {10}^{-7}\times 1500\times 5$

$B=3\pi \times {10}^{-3}T$

Now, the magnetic induction at one end on the axis of solenoid

$B=\frac{{\mu }_{0}Ni}{2}$

$B=\frac{4\pi \times {10}^{-7}\times 1500\times 5}{2}$

$B=1.5\pi \times {10}^{-3}T$

Hence, the magnetic induction inside of the solenoid is $3\pi \times {10}^{-3}T$ and at one end on the axis of solenoid is $1.5\pi \times {10}^{-3}T$