Question

A wire carrying a current is shaped in the form of a circular loop of radius 3.0mm If the magnetic field strength that this current produces at the center of the loop is 1.1mT, what is the magnitude of the current that flows through the wire? (μo = 4π x10^-7 T. m/A) A) 5.3A B) 16A C) 23 A D) 9.1A

Answer 1

Current, I = 5.3 A

Step-by-step explanation:

It is given that,

Radius of circular loop, r = 3 mm = 0.003 m

Magnetic field strength, B = 1.1 mT = 0.0011 T

We need to find the magnitude of the current that flows through the wire. The magnetic field for a current carrying wire is given by :

`B=(\mu_o I)/(2r)`

`I=(2Br)/(\mu_o)`

`I=(2* 0.0011\ T* 0.003\ m)/(4\pi* 10^(-7))`

I = 5.25 A

or

I = 5.3 A

So, the magnitude of the current that flows through the wire is 5.3 A. Hence, this is the required solution.