A thin flexible wire of length L is connected to a battery (negligible in size as compared to the wire) to form a closed loop. The battery drives a current I in the wire. When the system is put in a uniform magnetic field of strength B perpendicular to the plane in which the wire is placed, the wire takes the shape of a circle. The tension in the wire is

Question

A thin flexible wire of length L is connected to a battery (negligible in size as compared to the wire) to form a closed loop. The battery drives a current I in the wire. When the system is put in a uniform magnetic field of strength B perpendicular to the plane in which the wire is placed, the wire takes the shape of a circle. The tension in the wire is (A) IBL (B) (I B L/π ) (C) (I B L/2 π ) (D) (I B L/4 π )

Tardigrade Admin · Accepted Answer

L=2π R R=(L/2 π ) 2Tsin (d θ ) = Fm <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-a9kug2ujsbjsi2sp.jpg" /> For small angles, sin (. d θ .) ≈ d θ 2T(d θ )=I(d L)Bsin 90 ° =I (R . d θ ).B T=IRB=(I L B/2 π )

$I B L$

$\frac{I B L}{π}$

$\frac{I B L}{2 π}$

$\frac{I B L}{4 π}$

$L = 2 π R$
$R = \frac{L}{2 π}$
$2 T s in (d θ) = F_{m}$

For small angles, $s in (d θ) \approx d θ$
$2 T (d θ) = I (d L) B s in 9 0^{\circ}$
$= I (R . d θ) . B$
$T = I RB = \frac{I L B}{2 π}$

IBL

πIBL​

2πIBL​

4πIBL​

L=2πR R=2πL​ 2Tsin(dθ)=Fm​ For small angles, sin(dθ)≈dθ 2T(dθ)=I(dL)Bsin90∘ =I(R.dθ).B T=IRB=2πILB​

$I B L$

$\frac{I B L}{π}$

$\frac{I B L}{2 π}$

$\frac{I B L}{4 π}$

$L = 2 π R$
$R = \frac{L}{2 π}$
$2 T s in (d θ) = F_{m}$

For small angles, $s in (d θ) \approx d θ$
$2 T (d θ) = I (d L) B s in 9 0^{\circ}$
$= I (R . d θ) . B$
$T = I RB = \frac{I L B}{2 π}$