A conducting wire of uniform cross-section is bent to form an equilateral triangle ABC of side L . A current I enters at A and leaves at C . What will be the magnetic field at the centroid O of the triangle ?

Question

A conducting wire of uniform cross-section is bent to form an equilateral triangle ABC of side L . A current I enters at A and leaves at C . What will be the magnetic field at the centroid O of the triangle ? (A) 2√3(μ 0 I/4 π L) (B) (μ 0 I/4 π L) (C) √3(μ 0 I/4 π L) (D) Zero

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-gau3u9coshixkfz4.jpg" /> B at O:(i) textdue to IAB text is BAB=K⋅ (I/3) bigodot (.ii.) due to IBC is BBC=K(I/3) bigodot (.iii.) due to IAC is BAC=K× (2 I/3) bigotimes net B at O=K(I/3)+K(I/3)-K(2 I/3)=0 here K is any constant

Q. A conducting wire of uniform cross-section is bent to form an equilateral triangle $A BC$ of side $L$ . A current $I$ enters at $A$ and leaves at $C$ . What will be the magnetic field at the centroid $O$ of the triangle ?

$23 \frac{μ _{0} I}{4 π L}$

$\frac{μ _{0} I}{4 π L}$

$3 \frac{μ _{0} I}{4 π L}$

Zero

$B$ at $O : (i) due to I_{A B} is B_{A B} = K \cdot \frac{I}{3} ⨀$
$(ii)$ due to $I_{BC}$ is $B_{BC} = K \frac{I}{3} ⨀$
$(iii)$ due to $I_{A C}$ is $B_{A C} = K \times \frac{2 I}{3} ⨂$
net $B$ at $O = K \frac{I}{3} + K \frac{I}{3} - K \frac{2 I}{3} = 0$
here $K$ is any constant

Q. A conducting wire of uniform cross-section is bent to form an equilateral triangle ABC of side L . A current I enters at A and leaves at C . What will be the magnetic field at the centroid O of the triangle ?

23​4πLμ0​I​

4πLμ0​I​

3​4πLμ0​I​