Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needl

5.20(a). A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45 \degree with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.  Determine the horizontal component of the earth’s magnetic field at the location.

Answers (1)

best_answer

Given,

Number of turns in the coil, n = 30

Radius of coil, r = 12cm = 0.12m

Current in the coil, I = 0.35A

The angle of dip, \delta = 45o

We know, Magnetic fields due to current carrying coils, B = \mu_{0}nI/2r

B = 4\pi \times10^{-7}\times30\times0.35/2\times0.12

= 5.49 \times 10^{-5}T

Now, Horizontal component of the earth’s magnetic field = Bsin\delta

= 5.49 \times 10^{-5}T× sin45^o

=3.88\times10^{-5}T (Hint: Take sin45o as 0.7)

=0.388 G

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

CBSE Central Sector Scheme of Scholarship 2025 applications open on scholarships.gov.in
anam-khan

CBSE Board 2025: Class 10th, 12th supplementary date sheet out; exams from July 15
anam-khan

CBSE issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee

Latest Question