A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a
permanent magnet such that B is in plane of the coil. If due to current i in the triangle a torque T acts on it, the side l of the triangle is

$\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{\frac{1}{2}}$

$\frac{2}{\sqrt{3}} (\frac{τ}{Bi})$

$2 {(\frac{τ}{\sqrt{3} Bi})}^{\frac{1}{2}}$

$\frac{1}{\sqrt{3}} \frac{τ}{Bi}$

Solution

$2 {(\frac{τ}{\sqrt{3} Bi})}^{\frac{1}{2}}$

Torque acting on equilateral triangle in a magnetic field $\vec{B}$ is

$τ = i AB sinθ$

Area of triangle LMN

A = 1/2 × base × height

A = 1/2 ×l × sin 60^o

(since an equilateral triangle has an angle of 60^o each)

$A = \frac{\sqrt{3}}{2} l^{2} and θ = 90^{o}$

Substituting the given value in expression for torque, we have

$τ = i \times \frac{\sqrt{3}}{4} l^{2} B \sin 90^{o} τ = \frac{\sqrt{3}}{4} {il}^{2} B (∵ \sin 90^{o} = 1) Hence, l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{\frac{1}{2}}$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that B is in plane of the coil. If due to current i in the triangle a torque T acts on it, the side l of the triangle is 23τBi12 23τBi 2τ3Bi12 13τBi

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