Question

Wire AB carrying a current I1 is placed near another long wire CD
carrying current I2. If AB is free to move then it will move :
(A) Towards left
(B) Towards Right
(C) Upwards
(D) Downwards​

Answer 1

Given:

Wire AB carrying a current I1 is placed near another long wire CD carrying current I2.

To find:

Movement of AB.

Solution:

Wire CD will create a Magnetic Field intensity near AB in a direction in the plane of the paper.

$\vec{B} = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{2(I2)}{x} \bigg \} \: \: ( - \hat{k})$

Now, AB wire will experience a force upward due to the magnetic field intensity of CD.

$\therefore \: \vec{f} =( I1) \: ( \vec{L} \times \vec{B})$

$= > \: \vec{f} =( I1) \: \bigg \{ L \: (\hat{i} )\times \dfrac{ \mu_{0}}{4\pi} \dfrac{2(I2)}{x} \: \: ( - \hat{k}) \bigg \}$

$= > \: \vec{f} = \bigg \{ \dfrac{ \mu_{0}}{4\pi} \dfrac{2(I1)(I2)L}{x} \: \: ( \hat{j}) \bigg \}$

So, force will be experienced upwards.

Wire AB will move upwards (along y axis).