Question

A small circular flexible loop of wire of radius r carries a current i. it is placed in a uniform magnetic field

b. the tension in the loop will be doubled if

Answer 1

Let a small elementary length dl cut into circle as shown in figure, dθ is the angle subtended by dl into centre of circle . so, elementary length , dl = Rdθ
Now, F = i(dl × B)
= idl.Bsin90° [ because elementary length are perpendicular upon magnetic field ]
= BiRdθ

Also tension acts on just opposite side of force as you can see in figure .
So, at equilibrium condition ,
2Tsindθ/2 = BiRdθ
actually, dθ/2 is very small angle so, sindθ/2 ≈ dθ/2
Hence, 2Tdθ/2 = BiRdθ
T = BiR

So, you can see tension in loop is given by , T = BiR
Tension in loop will be doubled when (1) magnetic field will be doubled
(2) Current through the ring will be doubled
(3) Radius, R will be doubled.

Answer 2

The force acting on the small circular flexible loop of wire of radius r is given by :
F=mBSinθ

where m is given by πr²l

F=πr²l Bsinθ
So Tension force will be doubled when :
--->I is doubled
---> B is doubled .