jee-main 2019 Q19

jee-main · India · session2_10apr_shift1 Not Maths
Two wires $A$ \& $B$ are carrying currents $I _ { 1 }$ and $I _ { 2 }$ as shown in the figure. The separation between them is $d$. A third wire C carrying a current I is to be kept parallel to them at a distance $x$ from A such that the net force acting on it is zero. The possible values of $x$ are:
(1) $x = \pm \frac { \mathrm { I } _ { 1 } \mathrm {~d} } { \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) }$
(2) $x = \left( \frac { \mathrm { I } _ { 1 } } { \mathrm { I } _ { 1 } + \mathrm { I } _ { 2 } } \right) \mathrm { d }$ and $x = \frac { \mathrm { I } _ { 2 } } { \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) } \mathrm { d }$
(3) $x = \left( \frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } + \mathrm { I } _ { 2 } } \right) \mathrm { d }$ and $x = \left( \frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } } \right) \mathrm { d }$
(4) $x = \left( \frac { \mathrm { I } _ { 1 } } { \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } } \right) \mathrm { d }$ and $x = \frac { \mathrm { I } _ { 2 } } { \left( \mathrm { I } _ { 1 } + \mathrm { I } _ { 2 } \right) } \mathrm { d }$
Two wires $A$ \& $B$ are carrying currents $I _ { 1 }$ and $I _ { 2 }$ as shown in the figure. The separation between them is $d$. A third wire C carrying a current I is to be kept parallel to them at a distance $x$ from A such that the net force acting on it is zero. The possible values of $x$ are:\\
(1) $x = \pm \frac { \mathrm { I } _ { 1 } \mathrm {~d} } { \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) }$\\
(2) $x = \left( \frac { \mathrm { I } _ { 1 } } { \mathrm { I } _ { 1 } + \mathrm { I } _ { 2 } } \right) \mathrm { d }$ and $x = \frac { \mathrm { I } _ { 2 } } { \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) } \mathrm { d }$\\
(3) $x = \left( \frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } + \mathrm { I } _ { 2 } } \right) \mathrm { d }$ and $x = \left( \frac { \mathrm { I } _ { 2 } } { \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } } \right) \mathrm { d }$\\
(4) $x = \left( \frac { \mathrm { I } _ { 1 } } { \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } } \right) \mathrm { d }$ and $x = \frac { \mathrm { I } _ { 2 } } { \left( \mathrm { I } _ { 1 } + \mathrm { I } _ { 2 } \right) } \mathrm { d }$