Question

A coil in the shape of an equilateral triangle of sides I is suspended between the pole pieces of a permanent magnet such that B is in the plane of the coil. If due to a current ‘i’ in the triangle, a torque T acts on it, the side I of the triangle is:
A. $\dfrac{2}{\sqrt{3}}{{\left( \dfrac{T}{Bi} \right)}^{\dfrac{1}{2}}}$
B. $\dfrac{2}{3}\left( \dfrac{T}{Bi} \right)$
C. $2{{\left( \dfrac{T}{\sqrt{3}Bi} \right)}^{\dfrac{1}{2}}}$
D. $\dfrac{1}{\sqrt{2}}\left( \dfrac{T}{Bi} \right)$

Answer 1

Hint: As the coil is carrying current in the shape of an equilateral triangle and suspended between pieces of magnets, it exerts a torque. So, we shall use the torque equation by substituting the known values to find the answer.

Complete step by step solution:
A current carrying coil or loop placed in between magnetic fields experiences a torque. Let us now consider the general equation of torque which is given as
$\tau =rF\sin \theta$
Where, ‘F’ is the force, ‘r’ is the distance and ‘θ’ is the angle between them.
Each segment on the loop is perpendicular to magnetic field B. So, the force can be written as
$F=IlB$
On substituting ‘F’ in general equation, we get
$\tau =\omega IlB\sin \theta$
($r=\dfrac{\omega }{2}+\dfrac{\omega }{2}=\omega$ due to each vertical component)
If there are many loops with ‘N’ turns, we will get N times torque of one loop and $A=\omega l$. Therefore, the expression for torque becomes
$\tau =BiNA\sin \theta$
The current is flowing in the clockwise direction through the equilateral triangle and has a magnetic field in the direction of the z-axis.
The torque for the coil can be expressed as
$\tau =BiNA\sin \theta$
$\tau =BiNA\sin {{90}^{\circ }}$
Area of equilateral triangle is calculated as
$A=\dfrac{\sqrt{3}}{4}{{I}^{2}}$
Where ‘I’ is the value of sides. Let us substitute ‘A’ in the equation of torque and assume N=1,
$\tau =Bi\times \dfrac{\sqrt{3}}{4}{{I}^{2}}\times 1$
On rearranging,
${{I}^{2}}=\dfrac{4T}{\sqrt{3}Bi}$
We shall take square root on both sides, the equation becomes
$I=2{{\left( \dfrac{T}{Bi\sqrt{3}} \right)}^{\dfrac{1}{2}}}$
Therefore, the correct answer for the given question is option (C).

Note: The torque equation can be valid for loops of any shapes. The loop carrying a current I, has N turns, each of area A and perpendicular makes an angle θ with the magnetic field B. The net force on the loop will be zero.