वायु में रखे गए 100π A विद्युत धारा प्रवाहित करने वाले एक लंबे वृत्तीय चालक से 10 cm की दूरी पर चुंबकीय बल और अभिवाह घनत्व के संबंधित मानों की गणना कीजिए?

This question was previously asked in
SSC JE Electrical 9 Oct 2023 Shift 3 Official Paper-I
View all SSC JE EE Papers >
  1. 6.28 × 10-7 AT/m और 500 Wb/m2
  2. 50 Wb/m2 और 6.28 × 10-7 AT/m
  3. 500 AT/m और 6.28 × 10-4 Wb/m2
  4. 1500 Wb/m2 और 3.14 × 10-4 AT/m

Answer (Detailed Solution Below)

Option 3 : 500 AT/m और 6.28 × 10-4 Wb/m2
Free
Electrical Machine for All AE/JE EE Exams Mock Test
7.7 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

संकल्पना

एक वृत्तीय चालक के लिए चुंबकीय बल (H) निम्न द्वारा दिया जाता है?:

\(H={I\over 2π d}\)

जहाँ, I = धारा & d = दूरी

फ्लक्स घनत्व (B) निम्न द्वारा दिया गया है:

B = μoμr × H

μo = पारं पारगम्यता

μr = सापेक्ष पारगम्यता

गणना

दिया गया है,  I = 100π A

d = 10 cm

\(H={100π\over 2π 10× 10^{-2}}\)

H = 500 AT/m

वायु के लिए, μr = 1

B = 4π × 10-7 × 500 

B = 6.28 × 10-4 AT/m

Latest SSC JE EE Updates

Last updated on Jul 1, 2025

-> SSC JE Electrical 2025 Notification is released on June 30 for the post of Junior Engineer Electrical, Civil & Mechanical.

-> There are a total 1340 No of vacancies have been announced. Categtory wise vacancy distribution will be announced later.

-> 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.

More Magnetic Flux Density Questions

More Magnetostatics Questions

Get Free Access Now
Hot Links: online teen patti real money teen patti palace teen patti bindaas teen patti wink