A conducting rod of mass '\( m \)' and length '\( l \)' is hung horizontally with the help of two insulating threads such that the rod is along east-west direction. Now current \( I \) is set up from west to east direction. What should be magnitude and direction of magnetic field. So that tension in the threads becomes zero?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( B = \displaystyle\frac{mg}{Il} \), towards north
The tension becomes zero when the magnetic force on the rod balances its weight, so the force must be vertically upwards and equal to \( mg \). The force on a straight current-carrying rod is \( \vec{F} = I\,\vec{l}\times\vec{B} \).
With \( \vec{B} \) horizontal and perpendicular to the rod, \( F = IlB \):
\( \displaystyle IlB = mg \;\Rightarrow\; B = \frac{mg}{Il} \)
For the direction take east as \( \hat{\imath} \), north as \( \hat{\jmath} \) and vertically up as \( \hat{k} \). The current is from west to east, so \( \vec{l} = l\,\hat{\imath} \).
\( \hat{\imath}\times\hat{\jmath} = \hat{k}\ \text{(upwards)} \)
\( \hat{\imath}\times(-\hat{\jmath}) = -\hat{k}\ \text{(downwards)} \)
So the field must point towards north to give an upward force.