Question

A magnetic field of 14 x 10^-3 T exists at the corner of a circular coil of radius 20cm having 15

Answer 1

The current in the coil is 297 A

Explanation:

The magnetic field of a circular coil carrying a current is given by

$B=\frac{\mu_0 NI}{2R}$

where

$\mu_0 =4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

N is the number of turns in the coil

I is the current in the coil

R is the radius of the coil

For the coil in this problem, we have:

$B=14\cdot 10^{-3} T$

R = 20 cm = 0.20 m

N = 15

Solving the equation for I, we find the current in the coil:

$I=\frac{2BR}{\mu_0 N}=\frac{2(14\cdot 10^{-3})(0.20)}{(4\pi \cdot 10^{-7})(15)}=297 A$

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