Question

A straight long current carrying conductor and a finite current carrying rod of length $3a$ carrying same current are mutually perpendicular to each other as shown in figure.

• A. Net magnetic force on the rod is $\frac{\mu_0 I^2}{\pi}$ln (2)
• B. The distance of point of application of magnetic force on the rod from wire is $\frac{3a}{2\ln{(2)}}$
• C. If the system is released in free space with long wire being fixed, initially rod will move toward point $P$
• D. If the system is released in free space, initially both wire and rod move toward each other

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The problem involves calculating the magnetic force between a long straight current-carrying conductor and a finite current-carrying rod, and analyzing their subsequent motion.

1. Magnetic Field due to the Long Wire: The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ from it is given by: $B = \frac{\mu_0 I}{2\pi r}$

Assuming the long wire is along the y-axis and carries current $I$ in the +y direction (upwards). The rod is along the x-axis, carrying current $I$ in the +x direction (to the right).
Using the right-hand thumb rule, the magnetic field produced by the long wire at the location of the rod (for $x>0$) is directed into the page (in the -z direction). So, $\vec{B} = \frac{\mu_0 I}{2\pi r} (-\hat{k})$.

2. Net Magnetic Force on the Rod (Option A): Consider a small element $dr$ of the rod at a distance $r$ from the long wire. The current in this element is $I$, and its length vector is $d\vec{l} = dr \hat{i}$.
The force $d\vec{F}$ on this element is given by $d\vec{F} = I (d\vec{l} \times \vec{B})$.
$d\vec{F} = I (dr \hat{i} \times \frac{\mu_0 I}{2\pi r} (-\hat{k}))$
$d\vec{F} = \frac{\mu_0 I^2}{2\pi r} dr (\hat{i} \times (-\hat{k}))$

Since $\hat{i} \times (-\hat{k}) = -(\hat{i} \times \hat{k}) = -(-\hat{j}) = \hat{j}$, the force is in the +y direction (upwards).
The magnitude of the force is $dF = \frac{\mu_0 I^2}{2\pi r} dr$.
The rod extends from a distance $r=a$ to $r=4a$ from the long wire.
The total force $F$ on the rod is obtained by integrating $dF$ over the length of the rod:
$F = \int_{a}^{4a} \frac{\mu_0 I^2}{2\pi r} dr = \frac{\mu_0 I^2}{2\pi} \int_{a}^{4a} \frac{1}{r} dr$
$F = \frac{\mu_0 I^2}{2\pi} [\ln(r)]_{a}^{4a} = \frac{\mu_0 I^2}{2\pi} (\ln(4a) - \ln(a))$
$F = \frac{\mu_0 I^2}{2\pi} \ln\left(\frac{4a}{a}\right) = \frac{\mu_0 I^2}{2\pi} \ln(4)$

Since $\ln(4) = \ln(2^2) = 2\ln(2)$,
$F = \frac{\mu_0 I^2}{2\pi} (2\ln(2)) = \frac{\mu_0 I^2}{\pi} \ln(2)$
So, Option A is correct.

3. Distance of Point of Application of Magnetic Force (Option B): The point of application of the net force (center of force) $x_{eff}$ is found by setting the total torque equal to the torque produced by the net force acting at $x_{eff}$.
The torque $d\tau$ due to the elemental force $dF$ at distance $r$ from the long wire is $d\tau = r \cdot dF$.
$d\tau = r \left(\frac{\mu_0 I^2}{2\pi r}\right) dr = \frac{\mu_0 I^2}{2\pi} dr$
The total torque $\tau$ about the long wire is:
$\tau = \int_{a}^{4a} \frac{\mu_0 I^2}{2\pi} dr = \frac{\mu_0 I^2}{2\pi} [r]_{a}^{4a}$
$\tau = \frac{\mu_0 I^2}{2\pi} (4a - a) = \frac{\mu_0 I^2}{2\pi} (3a)$

The effective distance $x_{eff}$ is given by $\tau = F \cdot x_{eff}$, so $x_{eff} = \frac{\tau}{F}$.
$x_{eff} = \frac{\frac{\mu_0 I^2}{2\pi} (3a)}{\frac{\mu_0 I^2}{\pi} \ln(2)} = \frac{3a/2}{\ln(2)} = \frac{3a}{2\ln(2)}$
So, Option B is correct.

4. Initial Motion of the Rod (Long Wire Fixed) (Option C): As determined in step 2, the magnetic force on the rod is directed upwards (along +y direction), parallel to the long wire.
Point P is shown above the rod in the figure. Therefore, if the long wire is fixed, the rod will initially move upwards, towards point P.
So, Option C is correct.

5. Initial Motion of Both Wire and Rod (Released in Free Space) (Option D): According to Newton's third law, the force exerted by the long wire on the rod is equal in magnitude and opposite in direction to the force exerted by the rod on the long wire.
Since the force on the rod is upwards, the force on the long wire must be downwards.
Thus, if released in free space, the rod will move upwards, and the long wire will move downwards. Both will move parallel to each other, but in opposite directions. They will not move "toward each other" in the sense of reducing the perpendicular distance between them.
So, Option D is incorrect.

Final check of all options: A. Correct. B. Correct. C. Correct. D. Incorrect.