Question

Two solenoids are each made of 2 000 turns of copper wire per meter. solenoid 1 is 2 m long, while solenoid 2 is 1 m long. when equal currents are present in the two solenoids, the ratio of the magnetic field b1 along the axis of solenoid 1 to the magnetic field b2 along the axis of solenoid 2, b1/b2, is

Answer 1

The ratio of the magnetic field B₁ along the axis of solenoid 1 to the magnetic field B₂  along the axis of solenoid 2 is; B₁/B₂ = 1

What is the ratio of magnetic field?

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

B = μNI

where,

B is Magnetic Field due to solenoid

μ is permeability of free space

N is Number of turns per unit length

I is current passing through the solenoid

Thus, magnetic field for the first solenoid is;

B₁ = μN₁I₁

Similarly, magnetic field for the second solenoid is;

B₂ = μN₂I₂

Thus;

Ratio of B₁ to B₂ is;

B₁/B₂ = μN₁I₁/μN₂I₂

Since number of turns and current in both solenoids are the same, then we have; B₁/B₂ = 1

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