Question

Two solenoids of equal length are each made of 2000 turns of copper wire per meter. Solenoid I has a 5.00 cm radius; solenoid II a 10.0 cm radius. When equal currents are present in the two solenoids, the ratio of the magnitude of the magnetic field BIalong the axis of solenoid I to the magnitude of the magnetic field BIIalong the axis of solenoid II, BI/BII, is

Answer 1

BI/BII = 1

Step-by-step explanation:

The magnetic field due to a solenoid is given by the following formula:

`B = \mu nI\\`

where,

B = Magnetic Field due to solenoid

μ = permeability of free space

n = No. of turns per unit length

I = current passing through the solenoid

Now for the first solenoid:

`B_1 = \mu n_1I_1 \\`

For the second solenoid:

`B_2 = \mu n_2I_2\\`

Dividing both equations:

`(B_1)/(B_2) = (\mu n_1I_1)/(\mu n_2I_2)\\`

here, no. of turns and the current passing through each solenoid is same:

n₁ = n₂ and I₁ = I₂

Therefore,

`(B_1)/(B_2) = (\mu nI)/(\mu nI)\\`

BI/BII = 1