Question

A single-coil loop of radius r = 5.50 mm, shown below, is formed in the middle of an infinitely long, thin, insulated straight wire carrying the current i = 20.0 mA. What is the magnitude of the magnetic field at the center of the loop? Submit Answer Tries 0/99

Answer 1

Solution :

Given:

Radius of loop is (r) = 5.50 mm = 5.5 x 10^-3 m

Current (I) = 20 mA = 20 x 10^-3A

Since the magnetic field due to circular current-carrying conductor of radius r is: B_Loop= μ_o I / 2 r
and, The magnetic field due to the straight conducting wire at a distance r from the wire is: B_Wire = μ_o I / 2πr

Therefore, the net magnetic field at the center for the loop is: B = B_Loop + B_Wire
${\color{Golden} \therefore }$ B_center = (μ_oI / 2 r ) + ( μ_oI /2πr )
${\color{Golden} \therefore }$ B_center = ( μ_oI / 2r ) [ 1 + 1/π ]
${\color{Golden} \therefore }$ B_center = ( 1.318 ) ( μ_oI /2r )

${\color{Golden} \therefore }$ B_center = ( 1.318 ) [ ( 4π x 10^-7)(20 x 10^-3) / 2 (5.5 x 10^-3) ]

${\color{Golden} \therefore }$ B_center= 3.012 x 10^-6 T