Question

We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in solenoid 2 is three times that in solenoid 1. How does the field B2 at the center of solenoid 2 compare to B1 at the center of solenoid 1?

Answer 1

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:

$B=\mu_0\, \frac{N}{L} I$

Then, if we assign the subindex "1" to the quantities that define the magnetic field ($B_1$) inside solenoid 1, we have:

$B_1=\mu_0\, \frac{N_1}{L_1} I_1$

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 $L_2=L_1/2$

2) twice as many turns as solenoid 1. Then $N_2=2\,N_1$

3) three times the current of solenoid 1. Then $I_2=3\,I_1$

we obtain:

$B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1$