Question
In a long solenoid of 200 turns/cm carrying a current
i, the magnetic field at center is 6.28×10-2 T. If a second solenoid with 100 turns/cm carries current i/3, its field at center is:Solution
The magnetic field at the center of a long solenoid is given by the formula: B=μ0ni 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= μ0n 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= μ0n2i2 B2 = μ0(10000)(i/3) (Equation 2) To find B2, we can divide Equation 2 by Equation 1: 
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