In a long solenoid of 200 turns/cm carrying a current

Solution

The magnetic field at the center of a long solenoid is given by the formula: B=μ0​ni where: ·        B is the magnetic field strength ·        μ0 is the permeability of free space (4π×10^-7T ⋅ m/A) ·        n is the number of turns per unit length ·        i is the current flowing through the solenoid   For the first solenoid:
Number of turns per cm, n1=200turns/cm=200×100turns/m=20000turns/m Current, i1​=i Magnetic field at the center, B1​=6.28×10^-2T Using the formula: B1​= μ0​n i1​ 6.28×10^-2= μ0​(20000)i (Equation 1) For the second solenoid: Number of turns per cm, n2​=100turns/cm=100×100turns/m=10000turns/m Current, i1​=i/3 Magnetic field at the center, B2​ (to be found) Using the formula: B2​= μ0n2​i2​ B2​ = μ0​(10000)(i/3) (Equation 2) To find B2​, we can divide Equation 2 by Equation 1: