A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:
Question
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:
Solution
The magnetic field intensity (H) inside a solenoid can be calculated using Ampere's circuital law. The formula is:
H = Ni/L
where: N = number of turns in the solenoid, i = current flowing through the solenoid, and L = length of the solenoid.
Given: N = 1200 turns, i = 2 A, and L = 2 m,
we can substitute these values into the formula:
H = (1200 turns * 2 A) / 2 m = 1200 A/m.
So, the magnetic intensity at the center of the solenoid when a current of 2 A flows through it is 1200 A/m.
Similar Questions
Describe the magnetic field of a solenoid.
f the number of turns in a solenoid is increased, the strength of the magnetic field producedwill
A solenoid has core of a material with relative permeability 500 and its windings carry a current of 1 A. The number of turns of thesolenoid is 500 per metre. Then find magnetization of the material.
A closely wound solenoid of 800800 turns and area of cross-section 2.5×10−4 m22.5×10-4 m2 carries a current of 3.0 A3.0 A . What is its associated magnetic momen
Field inside a solenoid is A Directly proportional to the current flowing through the coil B Directly proportional to its length C Inversely porportional to total number of turns. D Inversely proportional to current.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.