JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 9)
Two wires A & B are carrying currents I1 & I2
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 :
_10th_April_Morning_Slot_en_9_1.png)
$$x = \left( {{{{I_1}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
$$x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
$$x = \left( {{{{I_1}} \over {{I_1} - {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d$$
$$x = \pm {{{I_1}d} \over {{I_1} - {I_2}}}$$
Explanation
Net force on wire carrying current I per unit
length is
$${{{\mu _0}{I_1}I} \over {2\pi x}} + {{{\mu _0}{I_2}I} \over {2\pi \left( {d - x} \right)}} = 0$$
$${{{I_1}} \over x} = {{{I_2}} \over {x - d}}$$
$$ \Rightarrow x = {{{I_1}d} \over {{I_1} - {I_2}}}$$
$${{{\mu _0}{I_1}I} \over {2\pi x}} + {{{\mu _0}{I_2}I} \over {2\pi \left( {d - x} \right)}} = 0$$
$${{{I_1}} \over x} = {{{I_2}} \over {x - d}}$$
$$ \Rightarrow x = {{{I_1}d} \over {{I_1} - {I_2}}}$$
Comments (0)
