A straight wire of length 0.5 m lies along the x-axis and carries a current of 2 A in the positive x-direction. It is placed in a uniform magnetic field \( \vec{B}=(0.3\,\hat{\jmath}+0.4\,\hat{k})\,\mathrm{T} \). The magnetic force on the wire is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( (-0.4\,\hat{\jmath}+0.3\,\hat{k})\,\mathrm{N} \)
The force on a straight conductor in a uniform field is \( \vec{F}=I\,\vec{l}\times\vec{B} \), where \( \vec{l} \) has the length of the wire and points along the current. Both components of this field are perpendicular to the wire, so each of them gives a force.
For this wire:
\( \begin{aligned} I\,\vec{l} &= 2\,\mathrm{A}\times0.5\,\hat{\imath}\,\mathrm{m} \\ &= 1.0\,\hat{\imath}\,\mathrm{A\,m} \end{aligned} \)
Using \( \hat{\imath}\times\hat{\jmath}=\hat{k} \) and \( \hat{\imath}\times\hat{k}=-\hat{\jmath} \), the force is:
\( \begin{aligned} \vec{F} &= 1.0\,\hat{\imath}\times(0.3\,\hat{\jmath}+0.4\,\hat{k})\,\mathrm{N} \\ &= (0.3\,\hat{k}-0.4\,\hat{\jmath})\,\mathrm{N} \end{aligned} \)
Hence, \( \vec{F}=(-0.4\,\hat{\jmath}+0.3\,\hat{k})\,\mathrm{N} \). This force has a magnitude of 0.5 N and acts at right angles to both the wire and the field.
NCERT (2026-27):
Magnetic Force on a Current-Carrying Conductor – “… Now nq vd is the current density j and |(nq vd)|A is the current I …. Thus,”
\( \vec{F}=I\,\vec{l}\times\vec{B} \)\( \qquad (4.4) \)
“where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector.”