Question 226: Magnetism and Matter

Question

A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 $ 	imes $ 10
-2
m. The magnetic field at the center of the triangle will be :

Accepted Answer

We know that magnetic field due to finite current carrying wire is
$B = {{{\mu _0}} \over {4\pi }}{I \over b}(\cos {	heta _1} + \cos {	heta _2})$ ..... (1)
Given that side of triangle = 4.5 $	imes$ 10
$-$2
m  = a; current = 1A since the triangle is equilateral, angle of each side will be 60$^\circ$.
Now, $	an 	heta  = {{Perpendicular} \over {Base}} \Rightarrow 	an 60^\circ  = {{a/2} \over b}$
$ \Rightarrow b = {a \over {2	an 60^\circ }} = {a \over {2\sqrt 3 }}$
Using equation (1), we get
$B = {{{\mu _0}} \over {4\pi }}{I \over {a/2\sqrt 3 }}(\cos 30^\circ  + \cos 30^\circ ) = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}2\cos 30^\circ $
$B = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}{{2 	imes \sqrt 3 } \over 2} = {{{\mu _0}} \over {4\pi }}{{6I} \over a}$
$ \Rightarrow B = {{{{10}^{ - 7}} 	imes 6 	imes 1} \over {4.5 	imes {{10}^{ - 2}}}} = 1.33 	imes {10^{ - 5}} \sim 2 	imes {10^{ - 5}}$ Wb/m
2

Answer

2 $ 	imes $ 10
-5
Wb/m
2

Answer

Zero

Answer

8 $ 	imes $ 10
-5
Wb/m
2

Answer

4 $ 	imes $ 10
-5
Wb/m
2

Magnetism and Matter

A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 \times 10 -2 m. The magnetic field at the center of the triangle will be :

Choose the correct answer: