A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle 2 θ0 at the centre of the circle (of which it forms an arch) then the tension in the wire is :

Question

A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle 2 θ0 at the centre of the circle (of which it forms an arch) then the tension in the wire is : (A) IBR (B) (IBR/ sinθ0) (C) (IBR/2 sinθ0) (D) (IBRθ0/ sinθ0)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/06029db4b2aee3bc24af668bde9c0412-.png" /> For equilibrium 2 T sin ((d θ/2))=Bi dl ⇒ 2 T((d θ/2))= Bi(R d θ) ⇒ T=Bi R= constant

$I BR$

$\frac{I BR}{s i n θ _{0}}$

$\frac{I BR}{2 s i n θ _{0}}$

$\frac{I BR θ _{0}}{s i n θ _{0}}$

For equilibrium
$2 T sin (\frac{d θ}{2}) = B i d l$
$\Rightarrow 2 T (\frac{d θ}{2}) = B i (R d θ)$
$\Rightarrow T = B i R =$ constant

IBR

sinθ0​IBR​

2sinθ0​IBR​

sinθ0​IBRθ0​​

For equilibrium 2Tsin(2dθ​)=Bidl ⇒2T(2dθ​)=Bi(Rdθ) ⇒T=BiR= constant

$I BR$

$\frac{I BR}{s i n θ _{0}}$

$\frac{I BR}{2 s i n θ _{0}}$

$\frac{I BR θ _{0}}{s i n θ _{0}}$

For equilibrium
$2 T sin (\frac{d θ}{2}) = B i d l$
$\Rightarrow 2 T (\frac{d θ}{2}) = B i (R d θ)$
$\Rightarrow T = B i R =$ constant