A square loop ABCD, carrying a current I2, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be

Question

A square loop ABCD, carrying a current I2, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be (A) (μ0I1I2/2π) (B) (μ0I1I2L/2π) (C) (2μ0I1I2L/3π) (D) (2μ0I1I2/3π)

Tardigrade Admin · Accepted Answer

F =1 LB sin θ B =(μ0 I /2 π r ) F AB =(- I 2 μ0 I 1 L /2 π x ) F CD =( I 2 μ0 1 L /2 π( L + x )) F =(μ0 I 1 L 2 l 2/2 π( x 2+1 x )) taking x =(1/2) 1 F =(2 μ0 I 1 I 2/3 π)

Q. A square loop ABCD, carrying a current I₂, is placed near and coplanar with a long straight conductor XY carrying a current I₁, as shown in figure. The net force on the loop will be

$\frac{μ _{0} I _{1} I _{2}}{2 π}$

$\frac{μ _{0} I _{1} I _{2} L}{2 π}$

$\frac{2 μ _{0} I _{1} I _{2} L}{3 π}$

$\frac{2 μ _{0} I _{1} I _{2}}{3 π}$

$F = 1 L B sin θ$
$B = \frac{μ _{0} I}{2 π r}$
$F_{A B} = \frac{- I _{2} μ _{0} I _{1} L}{2 π x}$
$F_{C D} = \frac{I _{2} μ _{0} 1 L}{2 π ( L + x )}$
$F = \frac{μ _{0} I _{1} L _{2} l ^{2}}{2 π ( x ^{2} + 1 x )}$ taking $x = \frac{1}{2} 1$
$F = \frac{2 μ _{0} I _{1} I _{2}}{3 π}$

Q. A square loop ABCD, carrying a current I2, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be

2πμ0​I1​I2​​

2πμ0​I1​I2​L​

3π2μ0​I1​I2​L​

3π2μ0​I1​I2​​