First slide

Magnetic dipole in a uniform magnetic field

Question

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

Difficult

A

$\frac{2}{\sqrt{3}} (\frac{τ}{Bi})$
B

$\frac{1}{\sqrt{3}} (\frac{τ}{Bi})$
C

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

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

Solution

The torque $τ$ is given by
$\begin{array}{l} τ = mBsin θ \\ ∴ τ = iABsin 90^{\circ} \\ or τ = iAB . . . . . . (1) \end{array} (∵ m = iA)$
where A = area of triangle.
To calculate A, consider fig.(8).
$\begin{array}{l} A = \frac{1}{2} \times Base \times height = \frac{1}{2} \times BC \times AD \\ Now AD = \sqrt{[l^{2} + (l / 2)^{2}]} = \frac{\sqrt{3}}{2} l \end{array} ∴ A = \frac{1}{2} \times l \times (\sqrt{3} / 2) l = (\sqrt{3} / 4) l^{2}$
Substituting the value of A in eq. (1), we get
$τ = i (\sqrt{3} / 4) l^{2} B \begin{array}{l} or l^{2} = \frac{τ}{(Bi)} \times (\frac{4}{\sqrt{3}}) \\ l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2} \end{array}$

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