Question

Two infinitely long wires each carrying the same current and pointing in \(+y\) direction are placed in the \(xy\)-plane, at \(x=-2\,cm\) and \(x=1\,cm\). An electron is fired with speed \(U\) from the origin making an angle of \(+45^\circ\) from the \(x\)-axis. The force on the electron at the instant it is fired is
\[ [B_0 \text{ is the magnitude of the field at origin due to the wire at } x=1\,cm \text{ alone}] \]

• A. \(\dfrac{-eUB_0}{2\sqrt{2}}(\hat{i}-\hat{j})\)
• B. \(\dfrac{-eUB_0}{2}(\hat{i}-\hat{j})\)
• C. \(\dfrac{-eUB_0}{\sqrt{2}}(\hat{i}-\hat{j})\)
• D. \(-eUB_0(\hat{i}-\hat{j})\)

Hint:

For a long straight current-carrying wire, use the right-hand rule to find magnetic field direction and use \[ B=\frac{\mu_0 I}{2\pi r} \] to compare field magnitudes.

Answer 1

The Correct Option is A

Step 1: Find the magnetic field direction due to the wire at \(x=1\,cm\).
Both wires carry current in \(+y\) direction.
For the wire at \(x=1\,cm\), the origin lies to the left of the wire.
Using the right-hand rule, magnetic field at the origin due to this wire is along \(+\hat{k}\).
Hence,
\[ \vec{B}_1=B_0\hat{k} \]

Step 2: Find the magnetic field due to the wire at \(x=-2\,cm\).
For the wire at \(x=-2\,cm\), the origin lies to the right of the wire.
Using the right-hand rule, magnetic field at the origin due to this wire is along \(-\hat{k}\).
The distance from this wire to the origin is \(2\,cm\).
Since magnetic field due to a long straight wire is inversely proportional to distance,
\[ B \propto \frac{1}{r} \] So, compared to the wire at \(x=1\,cm\),
\[ B_2=\frac{B_0}{2} \] Therefore,
\[ \vec{B}_2=-\frac{B_0}{2}\hat{k} \]

Step 3: Find the net magnetic field at the origin.
\[ \vec{B}_{net}=\vec{B}_1+\vec{B}_2 \] \[ \vec{B}_{net}=B_0\hat{k}-\frac{B_0}{2}\hat{k} \] \[ \vec{B}_{net}=\frac{B_0}{2}\hat{k} \]

Step 4: Write the velocity vector of the electron.
The electron is fired with speed \(U\) at \(45^\circ\) from the \(x\)-axis.
So,
\[ \vec{v}=U\cos45^\circ \hat{i}+U\sin45^\circ \hat{j} \] \[ \vec{v}=\frac{U}{\sqrt{2}}(\hat{i}+\hat{j}) \]

Step 5: Use magnetic force formula.
Magnetic force is given by
\[ \vec{F}=q(\vec{v}\times \vec{B}) \] For an electron,
\[ q=-e \] Thus,
\[ \vec{F}=-e\left[\frac{U}{\sqrt{2}}(\hat{i}+\hat{j})\times \frac{B_0}{2}\hat{k}\right] \] \[ \vec{F}=-e\frac{UB_0}{2\sqrt{2}}\left[(\hat{i}+\hat{j})\times \hat{k}\right] \] Now,
\[ \hat{i}\times\hat{k}=-\hat{j} \] and
\[ \hat{j}\times\hat{k}=\hat{i} \] So,
\[ (\hat{i}+\hat{j})\times\hat{k}=\hat{i}-\hat{j} \] Therefore,
\[ \vec{F}=-\frac{eUB_0}{2\sqrt{2}}(\hat{i}-\hat{j}) \]

Step 6: Final conclusion.
Hence, the force on the electron is
\[ \boxed{-\frac{eUB_0}{2\sqrt{2}}(\hat{i}-\hat{j})} \]