Question

A solenoid has 5000 turns and mean radius of 1m. It consists of a soft iron core with relative permeability of 3000. The magnitude of the magnetic field when current passing through the solenoid is 1 amp
1. 3 T
2. 2 T
3. 5 T
4. 1 T

Answer 1

Correct Answer - Option 1 : 3 T

CONCEPT:

Magnetic field strength:

• Consider a solenoid carrying n turns per unit length and carrying current I. The magnetic field strength in the interior of the solenoid, 

⇒ B0 = μ0nI

• The net field in the interior of the solenoid, 

⇒ B = B0 + B_m,

Where Bm is the field contributed by magnetic core given by, Bm = μ0M​

⇒ B = μ0(H + M)

• SI unit of magnetic field strength is weber/meter2 or Wb m-2. ​

EXPLAINATION:

Given: No of turns in a solenoid (N) = 5000, mean radius (r) = 1m, relative permeability (μ_r) = 3000, current passing through the solenoid (I) = 1 A

• Number of turns per unit length, 

\(⇒ n = \frac {5000}{2π r} =\frac{5000}{2\;× \;π\;× \;1}\)

• he magnetic field strength in the interior of the solenoid is given as, 

⇒ B0 = μnI

⇒ B  = μ₀μ_rni = 4π × 10^-7× 3000 × n × 1 =  3 Tesla

• Hence option 1) is correct.