Question

A square coil of edge l having n turns carries a current I, it is kept on a smooth horizontal plate. A uniform magnetic field B exists in direction parallel to an edge. The total mass of the coil is m. Find the minimum value of B for which the coil will start tipping over?
A. ${{\left( \dfrac{mg}{nil} \right)}^{2}}$
B. $\dfrac{2mg}{nil}$
C. $\dfrac{mg}{2nil}$
D. $\dfrac{mg}{nil}$

Answer 1

Hint: Magnetic field in the area tries to deflect coil from its position by exerting a deflecting torque to find the minimum value of it so the coil will start tipping we need to compare deflecting torque with restoring torque (which keeps the coil on the ground-acting in opposite direction).

Formula used:
To solve this problem formula of deflecting torque and restoring torque will be used, which is given by,
Deflecting torque=$niAB\sin (\theta )$
Restoring torque=$mg\dfrac{l}{2}$

Complete Step by step solution:
Given, length of edge = l, current=I, number of turns=n and mass =m
Magnetic field B to be calculated

Now, we know that deflecting torque is acting on the coil so,
Deflecting torque=$niAB\sin (\theta )$,
Here, $A={{l}^{2}}$(area of coil) and $\theta ={{90}^{\circ }}$

So, deflecting torque=$ni{{l}^{2}}B\sin (90)$
deflecting torque=$ni{{l}^{2}}B$
| :- sin(90)=1|

Now, restoring torque=force x (perpendicular distance from point of acting)=$mg\dfrac{l}{2}$
For tipping condition, restoring torque$\le $deflecting torque so,
$\Rightarrow mg\dfrac{l}{2}\le ni{{l}^{2}}B$
$\Rightarrow \dfrac{mg\dfrac{l}{2}}{ni{{l}^{2}}}\le B$
$\Rightarrow B=\dfrac{mg}{2nil}$

The minimum amount of magnetic field required so the coil will start tipping over is $B=\dfrac{mg}{2nil}$

So option (C) is correct.

Additional information:
As we know that coil is placed on a smooth surface, so as the magnetic field increases restoring torque is the only thing that stops the coil from tipping over because it acts outward in the direction opposite to deflecting torque.

Note:
Restoring torque is produced by weight of the body and acts on the center of mass, like in the above statements the perpendicular distance between center of mass and the point of contact for tipping is l/2. The oscillations of a simple pendulum will be a very good example for restoring torque as the center of gravity of the bob lies below the pivot.