A current of 1A is flowing on the sides of an equilateral triangle of sides 4.5× 10-2m. The magnetic field at the centroid of the triangle is

Question

A current of 1A is flowing on the sides of an equilateral triangle of sides 4.5× 10-2m. The magnetic field at the centroid of the triangle is (A)  2× 10-5T  (B)  4× 10-5T  (C)  8× 10-5T  (D)  1.2× 10-5T

Tardigrade Admin · Accepted Answer

B1=(μ 0I/4π a)( sin φ 1+ sin φ 2) tan 30° =(a/( (I)2 )) a=(I/2√3) B=(μ 0I/4π ( (I)2√3 ))( sin 60+ sin 60) B1=(6μ 0I/4π I) Btotal=3 B1 Btotal=4× 10-5T

Q. A current of 1A is flowing on the sides of an equilateral triangle of sides $4.5 \times 10^{- 2} m .$ The magnetic field at the centroid of the triangle is

$2 \times 10^{- 5} T$

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

$8 \times 10^{- 5} T$

$1.2 \times 10^{- 5} T$

$B_{1} = \frac{μ _{0} I}{4 πa} (sin ϕ_{1} + sin ϕ_{2})$
$tan 30^{\circ} = \frac{a}{( \frac{I}{2} )}$ $a = \frac{I}{2 3}$ $B = \frac{μ _{0} I}{4 π ( \frac{I}{2 3} )} (sin 60 + sin 60)$ $B_{1} = \frac{6 μ _{0} I}{4 π I}$ $B_{t o t a l} = 3 B_{1}$ $B_{t o t a l} = 4 \times 10^{- 5} T$

Q. A current of 1A is flowing on the sides of an equilateral triangle of sides 4.5×10−2m. The magnetic field at the centroid of the triangle is

2×10−5T

4×10−5T

8×10−5T

1.2×10−5T