COMEDK Exam 2019 Solved Paper

Given, radius of the coil =0.1m
No. of turns in the coil =10
Horizontal component of magnetic field,
BH=0.314×10−4T
The magnetic field at the centre of current carrying coil is given by the formula, B=‌

µ0nI

2R

According to question, magnetic field due to the coil obtains neutral point, in earth's magnetic field.
Hence, horizontal component of magnetic field,
‌BH=‌

µ0nI

2R

‌ Thus, ‌‌‌I=‌

2RBH

µ0n

⇒‌‌I=‌

2(0.1)(0.314×10−4)

10×4×(3.142)×10−7

=0.5A