A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, 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 two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is (A) IBL (B) (IBL/π) (C) (IBL/2 π) (D) (IBL/4 π)

Tardigrade Admin · Accepted Answer

L = 2 π R ∴ R= L / 2π 2T sin(d heta) = Fm For small angles, sin(d heta) ≈ d heta ∴ 2T (d heta) = I (dL) B sin 90° = I (2R ⋅ d heta) ⋅ B ∴ T = IRB = (ILB/2 π) ∴ Correct option is (c)

IBL

$\frac{I B L}{π}$

$\frac{I B L}{2 π}$

$\frac{I B L}{4 π}$

$L = 2 π R$
$∴ R = L /2 π$
$2 T s in (d h e t a) = F_{m}$
For small angles, $s in (d h e t a) \approx d h e t a$
$∴ 2 T (d h e t a) = I (d L) B s in 9 0^{\circ}$
$= I (2 R \cdot d h e t a) \cdot B$
$∴ T = I RB = \frac{I L B}{2 π}$
$∴$ Correct option is (c)

IBL

πIBL​

2πIBL​

4πIBL​

L=2πR ∴R=L/2π 2Tsin(dheta)=Fm​ For small angles, sin(dheta)≈dheta ∴2T(dheta)=I(dL)Bsin90∘ =I(2R⋅dheta)⋅B ∴T=IRB=2πILB​ ∴ Correct option is (c)

$\frac{I B L}{π}$

$\frac{I B L}{2 π}$

$\frac{I B L}{4 π}$

$L = 2 π R$
$∴ R = L /2 π$
$2 T s in (d h e t a) = F_{m}$
For small angles, $s in (d h e t a) \approx d h e t a$
$∴ 2 T (d h e t a) = I (d L) B s in 9 0^{\circ}$
$= I (2 R \cdot d h e t a) \cdot B$
$∴ T = I RB = \frac{I L B}{2 π}$
$∴$ Correct option is (c)