Question

if the number of turns becomes double but the length remain same then magnetic field in the solenoid becomes???

Answer 1

Gives 

solenoid

Case(1)

length

$l_1=l$

Turn

$N_1=n$

Magnetic field

B=B₁

case(2)

length

$l_2=l$

Turn

$N_2=2n$

Magnetic field

B=B₂

We know that

Magnetic field in solenoid

$B=\mu_0\frac{N}{l}i$

n=turn per unit length

we can written as 

$\frac{B_1}{B_2}=\frac{N_1}{N_2}\frac{l_2}{l_1} \rightarrow (1)$

Put value

$\frac{B_1}{B_2}=\frac{n}{2n}\frac{l}{l}$

$\frac{B_1}{B_2}=\frac{1}{2}$

$B_2=2B_1$

Final Magnetic field is equal to double of initial magnetic field