Given :
Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1.
To Find :
The ratio of the magnetic field in the interior of 2 to that in the interior of 1.
Solution :
We know, magnetic field in the interior of a solenoid is given by :
[tex]B = \dfrac{\mu ni}{L}[/tex]
Let, length of solenoid 2 is L.
Therefore, length of solenoid 1 is 6L.
[tex]B_a=\dfrac{\mu (6n)i}{L}\\\\B_b = \dfrac{\mu n i}{L}[/tex]
Dividing [tex]B_a[/tex] by [tex]B_b[/tex] :
[tex]\dfrac{B_a}{B_b}=\dfrac{\dfrac{\mu (6n)i}{L}}{\dfrac{\mu n i}{L}}\\\\\dfrac{B_a}{B_b}=6[/tex]
Therefore, the correct answer is C. 6.
