Question

A wire of mass \(100\,\text{g}\) is carrying a current of \(2\,\text{A}\) towards increasing \(x\) in the form \(y = x^2 \;(-2\,\text{m} \le x \le +2\,\text{m})\). This wire is placed in a magnetic field \(\vec{B} = -0.02\,\hat{k}\,\text{tesla}\). The acceleration of the wire (in \( \text{m s}^{-2} \)) is:

• A. \(-1.6\,\hat{j}\)
• B. \(-3.2\,\hat{j}\)
• C. \(1.6\,\hat{j}\)
• D. zero

Hint:

For curved current-carrying conductors in a uniform magnetic field: \[ \vec{F} = I \int d\vec{l} \times \vec{B} \] Odd functions over symmetric limits integrate to zero, simplifying force calculations.

Answer 1

The Correct Option is C

Step 1: Use the magnetic force on a current-carrying wire. The force on an element \(d\vec{l}\) is: \[ d\vec{F} = I \, (d\vec{l} \times \vec{B}) \]
Step 2: Write the differential length vector of the wire. The wire lies in the \(xy\)-plane with: \[ y = x^2 \Rightarrow \frac{dy}{dx} = 2x \] So, \[ d\vec{l} = dx\,\hat{i} + dy\,\hat{j} = ( \hat{i} + 2x\,\hat{j} )\,dx \]
Step 3: Evaluate the cross product \(d\vec{l} \times \vec{B}\). Given: \[ \vec{B} = -0.02\,\hat{k} \] \[ d\vec{l} \times \vec{B} = (\hat{i} + 2x\,\hat{j}) \times (-0.02\,\hat{k})\,dx \] Using vector products: \[ \hat{i} \times \hat{k} = -\hat{j}, \quad \hat{j} \times \hat{k} = \hat{i} \] \[ d\vec{l} \times \vec{B} = -0.02(-\hat{j} + 2x\,\hat{i})\,dx \] \[ = (0.02\,\hat{j} - 0.04x\,\hat{i})\,dx \]
Step 4: Integrate over the length of the wire. \[ \vec{F} = I \int_{-2}^{2} (0.02\,\hat{j} - 0.04x\,\hat{i})\,dx \] \[ \int_{-2}^{2} x\,dx = 0 \] \[ \vec{F} = I \left[ 0.02 \int_{-2}^{2} dx \right] \hat{j} \] \[ = I \,(0.02 \times 4)\,\hat{j} = 0.08 I\,\hat{j} \] With \(I = 2\,\text{A}\): \[ \vec{F} = 0.16\,\hat{j}\,\text{N} \]
Step 5: Find the acceleration of the wire. Mass: \[ m = 100\,\text{g} = 0.1\,\text{kg} \] \[ \vec{a} = \frac{\vec{F}}{m} = \frac{0.16}{0.1}\,\hat{j} = 1.6\,\hat{j}\,\text{m s}^{-2} \]
Hence, the acceleration of the wire is \(\boxed{1.6\,\hat{j}\,\text{m s}^{-2}}\).