# A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of – (i) an equilateral triangle of side a (ii) a square of side a The magnetic dipole moments of the coils in each case respectively are – I is the current through the coils. (a) 4Ia² and 3Ia² (b) √3Ia² and 3Ia² (c) 3Ia² and Ia² (d) 3Ia² and 4Ia²

By BYJU'S Exam Prep

Updated on: September 13th, 2023

Step I – It is given

12a is the length of uniform conducting wire

R is the resistance of a wire

I is the current through the coils

Step II – Formula to be used

The magnetic dipole moment of the coil is written as

µ = NiA

Where µ is the magnetic moment,

N is the total number of turns in the wire

i is the current in the coil

A is the area of cross-section

We know that

N = Total length of the wire/ Length of each turn

Step III – To calculate the magnetic dipole moment when the current carrying coil is an equilateral triangle shape

a is the side of an equilateral triangle

Length of equilateral triangular turn = a + a + a = 3a

Number of total triangular turns in the wire N = 12a/3a = 4

Area of equilateral triangle A = √3/4 x side2

Magnetic dipole moment µ = 4 x I x √3/4 x a2 = √3Ia2

Step IV – To calculate the magnetic dipole moment when the current carrying coil is a square shape

a is the side of the square

Length of square coil = a + a + a + a = 4a

Number of total square turns in the wire N = 12a/4a = 3

Area of square A = side2

Magnetic dipole moment µ = 3 x I x a2 = 3Ia2

Therefore, the magnetic moment in the first case is √3Ia2 and the magnetic moment in the second case is 3Ia2.

Summary:

## (d) 3Ia2 and 4Ia2

A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of

(i) an equilateral triangle of side a

(ii) a square of side a

The magnetic dipole moments of the coils in each case respectively are –

I is the current through the coils.

(b) √3la2 and 3Ia2.

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