Question

15. A rectangular loop ABCD is placed near on infinitelength current carrying wire. Magnetic force onthe loop is :25A15A10cm« 5cmi 25cm(source)RAMAASOOS​

Answer 1

Follow Right Hand Thumb Rule

Explanation:

• The sides AB and CD are symmetrically placed but having opposite direction of current.  

• Net force on AB and CD will be zero.  

• Net force acting on rectangle ABCD will be same as net force acting on AD and BC.

• Magnetic field due to infinite wire = $\frac {\mu_oI}{2\pi r}$ (Right Hand Thumb Rule for Direction)

I = 25 A

Magnetic field due to infinite wire at AD (r = 5 cm = 0.05 m), $B_a_t_A_D = \frac {\mu_o\times25}{2\pi 0.05}$ (inside paper)

Magnetic field due to infinite wire at BC (r = 3 cm = 0.03 m), $B_a_t_B_C = \frac {\mu_o\times25}{2\pi 0.03}$ (inside paper)

Force acting on a current carrying conductor is given by, F = $I(\overrightarrow{l} \times \overrightarrow{B})$

Here I = current flowing through the current carrying conductor = 15A, $\left |\overrightarrow{l} \right | = 10 cm$ = 0.1 m

(direction is in the direction of current),

Force acting on AD, $F_A_D = 15\times 0.1 \times B_a_t_A_D$ (towards left)

Force acting on BC, $F_B_C = 15\times 0.1 \times B_a_t_C_D$(towards right)

$Net Force = F_A_D - F_B_C$  (towards left)

= $15\times 0.1 \times B_a_t_A_D - 15\times 0.1 \times B_a_t_C_D$ (towards left)

= $15\times 0.1 (B_a_t_A_D - B_a_t_C_D)$ (towards left)

= $15\times 0.1 (\frac {\mu_o\times25}{2\pi 0.05} - \frac {\mu_o\times25}{2\pi 0.03})$ (towards left)

= $15\times 0.1 \times \frac {\mu_o\times25}{2\pi} (\frac {1}{0.05} - \frac {1}{0.3})$ (towards left)

= $15\times 0.1 \times \frac {\mu_o\times25}{2\pi} \times 16.667$ (towards left)

= $15\times 0.1 \times \frac {4\pi\times 10^-^7\times25}{2\pi} \times 16.667$ (towards left)

= $1.25 \times 10^-^4$ N (towards left i.e. attraction occurs)