Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A short bar magnet of magnetic movement 5.25 10 2 J T minus 1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on its normal

A short bar magnet of magnetic movement   5.25 \times 10 ^{-2} JT^{-1}  is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on  its normal bisector. 

Answers (1)

best_answer

Magnetic moment M is   = 5.25\times 10^{-2}\ JT^{-1}

Earth's field = 0.42 G.

The magnetic field due to magnet at equatorial line is given by: 

                                              B\ =\ \frac{\mu _0}{4\Pi }. \frac{M}{r^3}

or                                                  \frac{\mu _0}{4\Pi }. \frac{M}{r^3}\ =\ 0.42\times 10^{-4}

or                                                    r^3\ =\ 125\times 10^{-8}\ m

                                                         r\ =\ 0.05 m

     

 

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question