Question
Download Solution PDFDetermine the force required to separate two magnetic surfaces with a contact area of 4π cm2 and the magnetic flux density across the surface is 1 wb/m2.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The force required to separate two magnetic surfaces is given by:
\(F={B^2A\over 2μ_o}\)
where, B = Magnetic flux density
A = Contact area
μo = Permittivity of free space
Calculation
Given, A = 4π cm2 = 4π × 10-4 m2
B = 1 wb/m2
μo = 4π × 10-4
\(F={(1)^2\times 4 π × 10^{ −4 } \over 2\times 4\pi\times 10^{-7}}\)
F = 500 N
Last updated on May 29, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.
-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31.
-> Candidates with a degree/diploma in engineering are eligible for this post.
-> The selection process includes Paper I and Paper II online exams, followed by document verification.
-> Prepare for the exam using SSC JE EE Previous Year Papers.