Answer:
the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.
Explanation:
Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:
[tex]B=\mu_0\, \frac{N}{L} I[/tex]
Then, if we assign the subindex "1" to the quantities that define the magnetic field ([tex]B_1[/tex]) inside solenoid 1, we have:
[tex]B_1=\mu_0\, \frac{N_1}{L_1} I_1[/tex]
notice that there is no dependence on the diameter of the solenoid for this formula.
Now, if we write a similar formula for solenoid 2, given that it has :
1) half the length of solenoid 1 . Then [tex]L_2=L_1/2[/tex]
2) twice as many turns as solenoid 1. Then [tex]N_2=2\,N_1[/tex]
3) three times the current of solenoid 1. Then [tex]I_2=3\,I_1[/tex]
we obtain:
[tex]B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1[/tex]
