An equilateral triangle is made by uniform wires AB, BC, CA. A current I enters at A and leaves from the mid point of BC. If the lengths of each side of the triangle is L, the magnetic field B at the centroid O of the triangle is

Question

An equilateral triangle is made by uniform wires AB, BC, CA. A current I enters at A and leaves from the mid point of BC. If the lengths of each side of the triangle is L, the magnetic field B at the centroid O of the triangle is (A) (μ0/4π)((4I/L)) (B) (μ0/3π)((4I/L)) (C) (μ0/4π)((2I/L)) (D) Zero

Tardigrade Admin · Accepted Answer

The given situation can be shown as <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/acdc24ab5eb0141bba0fc7f7560f2b7d-.png" /> The magnetic field at centroid O= Magnetic field due to left part + Magnetic field due to right part ∴ B = B 1+ B 2=0 (∵ direction of B due to both parts is different of O )

Q. An equilateral triangle is made by uniform wires $A B, BC, C A . A$ current $I$ enters at $A$ and leaves from the mid point of $BC$ . If the lengths of each side of the triangle is $L$ , the magnetic field $B$ at the centroid $O$ of the triangle is

$\frac{μ _{0}}{4 π} (\frac{4 I}{L})$

$\frac{μ _{0}}{3 π} (\frac{4 I}{L})$

$\frac{μ _{0}}{4 π} (\frac{2 I}{L})$

Zero

The given situation can be shown as

The magnetic field at centroid $O =$ Magnetic field due to left part + Magnetic field due to right part $∴ B = B_{1} + B_{2} = 0$
$(∵$ direction of B due to both parts is different of $O$ )

Q. An equilateral triangle is made by uniform wires AB,BC,CA.A current I enters at A and leaves from the mid point of BC. If the lengths of each side of the triangle is L, the magnetic field B at the centroid O of the triangle is

4πμ0​​(L4I​)

3πμ0​​(L4I​)

4πμ0​​(L2I​)