A horizontal straight wire of mass m hangs from two light vertical threads in a uniform horizontal magnetic field that is perpendicular to the wire. When a current I is passed through the wire, the tension in each thread becomes zero. If the direction of the current is now reversed, keeping its magnitude the same, the total tension in the two threads will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2mg \)
The magnetic force on the wire, \( \vec{F}=I\,\vec{l}\times\vec{B} \), is vertical here because the field is horizontal and perpendicular to the wire. Reversing the current reverses \( \vec{l} \), so the force keeps its magnitude \( IlB \) but turns the other way.
With the first current the threads go slack, so the magnetic force acts upwards and balances the weight, \( IlB=mg \).
With the current reversed, the same force acts downwards, and the threads now hold up the weight and this force together.
The total tension in the two threads is then:
\( \begin{aligned} T &= mg+IlB \\ &= mg+mg \\ &= 2mg \end{aligned} \)