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: Field of the near wire at the origin.
Both wires carry current along $+y$. For the wire at $x = 1\ \text{cm}$, the right-hand rule puts its field at the origin along $+\hat{k}$ with magnitude $B_0$, so \[ \vec{B}_1 = B_0\hat{k}. \]
Step 2: Field of the far wire at the origin.
The wire at $x = -2\ \text{cm}$ has the origin on its other side, so its field points along $-\hat{k}$. Since $B \propto 1/r$ and this wire is twice as far, its magnitude is $B_0/2$: \[ \vec{B}_2 = -\frac{B_0}{2}\hat{k}. \]
Step 3: Net magnetic field.
\[ \vec{B} = B_0\hat{k} - \frac{B_0}{2}\hat{k} = \frac{B_0}{2}\hat{k}. \]
Step 4: Velocity of the electron.
Fired at $45^\circ$ from the $x$-axis with speed $U$, \[ \vec{v} = \frac{U}{\sqrt2}(\hat{i} + \hat{j}). \]
Step 5: Apply the Lorentz force with charge $-e$.
\[ \vec{F} = (-e)\,\vec{v}\times\vec{B} = -e\,\frac{U}{\sqrt2}(\hat{i}+\hat{j})\times\frac{B_0}{2}\hat{k}. \] Using $\hat{i}\times\hat{k} = -\hat{j}$ and $\hat{j}\times\hat{k} = \hat{i}$, \[ (\hat{i}+\hat{j})\times\hat{k} = -\hat{j} + \hat{i} = \hat{i} - \hat{j}. \]
Step 6: Assemble and conclude.
\[ \vec{F} = -e\cdot\frac{U}{\sqrt2}\cdot\frac{B_0}{2}(\hat{i}-\hat{j}) = \frac{-eUB_0}{2\sqrt2}(\hat{i}-\hat{j}). \]
\[ \boxed{\dfrac{-eUB_0}{2\sqrt2}(\hat{i}-\hat{j})} \]