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 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

15 (b). 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 axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Answers (1)

best_answer

The magnetic field at a distance R from the centre of the magnet on its axis:

B = \mu_{0}m/2\pi R^{3}

When the resultant field is inclined at 45° with earth’s field, B = H 

R^3 = \mu_{0}m/2\pi B = 4\pi\times10^{-7}\times5.25\times10^{-2}/2\pi\times0.42\times10^{-4}= 25 \times 10^{-5}m^3

Therefore, R = 0.063 m  = 6.3 cm 

Posted by

HARSH KANKARIA

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question