COMEDK Exam 2018 Solved Paper

I1=20A,I2=20A
‌r1=2‌cm
‌r2=(10+2)‌cm=12‌cm
The given situation is shown in the figure.
The forces on arm SR and PQ are equal in magnitude and opposite in direction and also having same line of action, hence they cancel to each other.
∴ Resultant force on the current loop PQRS,
‌F=FPS−FQR
‌‌=‌

µ0

2π

⋅‌

I1I2×l

r1

−‌

µ0

2π

⋅‌

I1I2l

r2

‌‌=‌

µ0

2π

⋅I1I2[[‌

1

r1

−‌

1

r2

].
‌‌=2×10−7×20×20×15×10−2[‌

1

2×10−2

−‌

1

12×10−2

]
‌‌=5×10−4N