Question

A circular coil of 120 turns and radius 0.18 m and carries current of 1a what is the manget field strength at a point on the axis of the coil at a distance of 0.18 m from centre of the coil

Answer 1

PLEASE MAKE ME BRAINEST

The magnitude of the magnetic field BB at the center of a circular coil having NN number of turns with radius RR and carrying a current II is given by

B=μ0NI2RB=μ0NI2R

Here, we have

B=M⋅10−3B=M⋅10−3 TT, I=1.5I=1.5 AA, R=1.5R=1.5 cm=1.5⋅10−2cm=1.5⋅10−2 mm, N=25N=25

Hence,

M⋅10−3T=(4π⋅10−7TmA)⋅25⋅(1.5A)2⋅(1.5⋅10−2m)M⋅10−3T=(4π⋅10−7TmA)⋅25⋅(1.5A)2⋅(1.5⋅10−2m)

⇒M=1.57⇒M=1.57 (Unit-less)

Answer 2

PLEASE MARK ME AS BRAINLIST ❤️