Two long straight wires P and Q carrying equal current 10 A ench were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1 If distance between wires is halved and currents on them are doubled, force F2 on 10 cm length of wire P will be:

Question

Two long straight wires P and Q carrying equal current 10 A ench were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1 If distance between wires is halved and currents on them are doubled, force F2 on 10 cm length of wire P will be: (A) 10 F 1 (B) 8 F 1 (C) (F1/8) (D) (F1/10)

Tardigrade Admin · Accepted Answer

Force per unit length between two parallel straight text wires =(μ0 i1 i2/2 π d) (F1/F2)=((μ0(10)2/2 π(5 cm ))/(μ0(20)2)2 π((5 cm )2)=(1/8) ⇒ F2=8 F1

$10 F_{1}$

$8 F_{1}$

$\frac{F _{1}}{8}$

$\frac{F _{1}}{10}$

Force per unit length between two parallel straight
$wires = \frac{μ _{0} i _{1} i _{2}}{2 π d}$
$\frac{F _{1}}{F _{2}} = \frac{\frac{μ _{0} ( 10 ) ^{2}}{2 π ( 5 c m )}}{\frac{μ _{0} ( 20 ) ^{2}}{2 π ( \frac{5 c m}{2} )}} = \frac{1}{8}$
$\Rightarrow F_{2} = 8 F_{1}$

10F1​

8F1​

8F1​​

10F1​​

Force per unit length between two parallel straight wires =2πdμ0​i1​i2​​ F2​F1​​=2π(25cm​)μ0​(20)2​2π(5cm)μ0​(10)2​​=81​ ⇒F2​=8F1​

$10 F_{1}$

$8 F_{1}$

$\frac{F _{1}}{8}$

$\frac{F _{1}}{10}$

Force per unit length between two parallel straight
$wires = \frac{μ _{0} i _{1} i _{2}}{2 π d}$
$\frac{F _{1}}{F _{2}} = \frac{\frac{μ _{0} ( 10 ) ^{2}}{2 π ( 5 c m )}}{\frac{μ _{0} ( 20 ) ^{2}}{2 π ( \frac{5 c m}{2} )}} = \frac{1}{8}$
$\Rightarrow F_{2} = 8 F_{1}$