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 is
4×10−5T
40T
0.4×105T
4×!0−2T
A
4×10−5T
B
0.4×105T
C
4×!0−2T
D
40T
Open in App
Solution
Verified by Toppr
Magnetic field at the centroid is given by,
B=3μ0I4πa(sinθ1+sinθ2)
Here, θ1=θ2=60
Length =4.5×10−2m
a=12cot60
=4.5×10−22√3
Therefore,
B=3×10−7×1×2√34.5×10−22√32=4×10−5T
B=4×10−5T
Was this answer helpful?
10
Similar Questions
Q1
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 is
View Solution
Q2
A current 1A is flowing in the sides of equilateral triangle of side 4.5×10−2m .The magnetic field at centroid of the triangle is
View Solution
Q3
A current of 1A is flowing along the sides of an equilateral triangle of side 4.5×10−2m. The magnetic field at the centroid of the triangle is (μ0=4π×10−7H/m)
View Solution
Q4
A current of 1amp is flowing in the sides of an equilateral triangle of side 4.5×10−2m. Find the magnetic field at the centroid of triangle.
View Solution
Q5
A current of 1A is flowing on the sides of an equilateral triangle of side 4.5×10−2m. The magnetic field at the centre of the triangle will be: