A uniform magnetic field \( \vec{B} \) is established along the positive z-direction. A rectangular loop of sides 'a' and 'b' carries a current of I as shown in figure. The torque in the loop is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( IabB\,(-\hat{\jmath}) \)
The torque on a current loop in a uniform field is \( \vec{\tau} = \vec{m}\times\vec{B} \), where \( \vec{m} = I\vec{A} \) and the direction of the area vector follows the right-hand rule from the sense of the current.
From the figure, the loop lies in the y-z plane: the current goes from O to C (\( +y \)), up the side at C (\( +z \)), back along the top (\( -y \)) and down the z-axis to O. Curling the fingers of the right hand in this sense, the thumb points along \( +x \).
\( \vec{A} = ab\,\hat{\imath} \)
\( \vec{m} = Iab\,\hat{\imath} \)
\( \vec{B} = B\,\hat{k} \)
\( \displaystyle \vec{\tau} = Iab\,\hat{\imath} \times B\,\hat{k} = IabB\,(\hat{\imath}\times\hat{k}) = IabB\,(-\hat{\jmath}) \)
using \( \hat{\imath}\times\hat{k} = -\hat{\jmath} \).