Question

A part of a wire carrying \(2.0\) A current and bent at \(90^\circ\) at two points is placed in a region of uniform magnetic field \[ \vec B=-0.50\,\hat k\,T. \]

Calculate the magnitude of the net force acting on the wire.

Answer 1

Force on the Bent Wire

Step 1: Use the property of uniform magnetic field. For a wire carrying current in a uniform magnetic field, \[ \vec F = I(\vec L\times\vec B), \] where \(\vec L\) is the displacement vector joining the initial and final points.

Step 2: Determine the effective length vector. From the figure, horizontal displacement \[ =50\,\text{cm}=0.50\,m. \] Vertical displacement \[ =-20\,\text{cm}=-0.20\,m. \] Therefore \[ \vec L = 0.50\hat i-0.20\hat j. \]

Step 3: Calculate \(\vec L\times\vec B\). Given \[ \vec B=-0.50\hat k. \] Hence \[ \vec L\times\vec B = (0.50\hat i-0.20\hat j)\times(-0.50\hat k). \] Using \[ \hat i\times\hat k=-\hat j, \qquad \hat j\times\hat k=\hat i, \] \[ \vec L\times\vec B = 0.25\hat j+0.10\hat i. \]

Step 4: Calculate force vector. \[ \vec F = I(\vec L\times\vec B). \] Since \[ I=2.0\,A, \] \[ \vec F = 0.20\hat i+0.50\hat j. \]

Step 5: Find magnitude. \[ F = \sqrt{(0.20)^2+(0.50)^2}. \] \[ F = \sqrt{0.29}. \] \[ F = 0.538\,N. \] Therefore, \[ \boxed{ F\approx0.54\,N } \]