A straight rod of length 2 m carrying a current of 1.4 A is placed at \( 45^\circ \) to a uniform magnetic field of 0.5 T. What is the magnitude of the force on the rod?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 N
A straight rod of length \( l \) carrying a current \( I \) in a uniform magnetic field feels the force \( \vec{F}=I\,\vec{l}\times\vec{B} \). Its magnitude is \( IlB\sin\theta \), where \( \theta \) is the angle between the rod and the field.
\( \begin{aligned} F &= IlB\sin\theta \\ &= 1.4\,\mathrm{A}\times2\,\mathrm{m}\times0.5\,\mathrm{T}\times\sin45^\circ \\ &= 1.4\times0.707\,\mathrm{N} \\ &\approx 0.99\,\mathrm{N} \\ &\approx 1\,\mathrm{N} \end{aligned} \)