Question

A long solenoid carrying a current of 2.0 A has 1000 turns per metre. If magnetic field inside the solenoid is 8π × 10^–k T, then find k

Answer 1

Explanation:

A solenoid is 2 m long and 3 cm in diameter. It has 5 layers of windings of 1000 turns each and carries a current of 5 A. What is the magnetic field at its center? Use the standard value of μ0

B=μ0ni = μ0lni

    = (4π×10−7)11000×5

B=2π×10−3    

force on the electron= qVB

      = (1.6×10−19)×104×2π×10−3N

      Fe=1.044×10−19N

Hence the force experienced by this electron is Fe=1.044×10−19N

Answer 2

Answer:

The value of k will be equal to 4 and magnetic field becomes 8π×10⁻⁴T.

Explanation:

A solenoid is a tightly wound helical loop from a insulated wire, such that its length is very large compared to its area.

Given, the current  passed through the solenoid, $I=2A$

The solenoid has number of turns, n = 1000 per meter.

The magnetic field inside the solenoid $B = 8 \pi * 10^{-k}T$

The magnetic field produced due to solenoid is:

$B= \mu_{o}nI$

Where  μ₀ is absolute permittivity of the free space or vacuum.

$\mu_{o}=4\pi *10^{-7}TmA^{-1}$

Put the value of B, I, n and  μ₀ in equation (1);

$8 \pi * 10^{-k}T=(4\pi *10^{-7}TmA^{-1})*(1000m^{-1})*(2A)$

$10^{-k}=10^{-7}*10^{3}$

$10^{-k}=10^{-4}$

$k=4$

Therefore, the value of k is equal to 4.