A current 1A is flowing in the sides of equilateral triangle of side 4-5times -0-2-m- The magnetic induction centroid of the triangle is4times -10-5-T40T0-4times -10-5-T4times -0-2-T

Question

Toppr Admin · Accepted Answer

Magnetic field at the centroid is given by,

$B = 3 \frac{μ_{0} I}{4 π a} (s i n θ_{1} + s i n θ_{2})$

Here, $θ_{1} = θ_{2} = 60$

Length $= 4.5 \times 10^{- 2} m$

$a = \frac{1}{2} c o t 60$

$= \frac{4.5 \times 10^{- 2}}{2 \sqrt{3}}$

Therefore,

$B = \frac{3 \times 10^{- 7} \times 1 \times 2 \sqrt{3}}{4.5 \times 10^{- 2}} \frac{2 \sqrt{3}}{2} = 4 \times 10^{- 5} T$

$B = 4 \times 10^{- 5} T$

A current 1A is flowing in the sides of equilateral triangle of side 4.5×!0−2m. The magnetic induction at centroid of the triangle is4×10−5T40T0.4×105T4×!0−2T

Magnetic field at the centroid is given by,B=3μ0I4πa(sinθ1+sinθ2)Here, θ1=θ2=60Length =4.5×10−2ma=12cot60=4.5×10−22√3Therefore, B=3×10−7×1×2√34.5×10−22√32=4×10−5TB=4×10−5T

A current $1 A$ is flowing in the sides of equilateral triangle of side $4.5 \times {! 0}^{- 2} m$ . The magnetic induction at centroid of the triangle is

$4 \times 10^{- 5} T$
$40 T$
$0.4 \times 10^{5} T$
$4 \times {! 0}^{- 2} T$