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

Answer 1

The correct option is A $2\pi \times {10}^{-3}T,\pi \times {10}^{-3}T$

Given: Length of solenoid = 1.5 m

Radius of solenoid= 2 cm

Number of turns= 10 turns/cm or 1000 turns/m

Current= 5A

Solution: (a) The magnetic field inside the solenoid is given by,

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

Substituting the values we get,

$B=4\pi \times {10}^{-7}\times 1000\times 5$

$B=2\pi \times {10}^{-3}T$

(b) The magnetic field at one end on the axis of the solenoid is half the value of the magnetic field inside the solenoid. So,

$B=\frac{\mu ni}{2}$

So we get,

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

Hence, the correct option is (A).