Question

A uniform magnetic field $\overrightarrow{B}$ fills a cylindrical volumes of radius r. A conducting ring of same radius (r) and total resistance R exactly fits into the region of magnetic field as shown. A perfect conducting rod (zero resistance) is being moved with constant velocity v across the magnetic field touches the ring as shown in the figure. The magnitude of force applied by external force when the rod is at a distance $\frac{r}{2}$ is

F=$\frac{NB^2r^2v}{R}$. Find the value of N.

Answer 1

27/2

Answer 2

The problem involves calculating the motional EMF induced in a conducting rod moving in a magnetic field, the equivalent resistance of the circuit formed by the rod and the ring, the current in the circuit, and finally the magnetic force on the rod, which must be balanced by the external force for constant velocity.

1. Length of the conducting rod ($L$) in the magnetic field: The rod is at a distance $y = r/2$ from the center of the ring. The equation of the circular ring is $x^2 + y^2 = r^2$. Substituting $y = r/2$ into the equation: $x^2 + \left(\frac{r}{2}\right)^2 = r^2$ $x^2 + \frac{r^2}{4} = r^2$ $x^2 = r^2 - \frac{r^2}{4} = \frac{3r^2}{4}$ $x = \pm \frac{\sqrt{3}}{2}r$ The length of the rod within the magnetic field (and touching the ring) is the distance between these two x-coordinates: $L = \frac{\sqrt{3}}{2}r - \left(-\frac{\sqrt{3}}{2}r\right) = \sqrt{3}r$

2. Induced electromotive force ($\mathcal{E}$): The rod moves with velocity $\overrightarrow{v}$ perpendicular to the uniform magnetic field $\overrightarrow{B}$ and its length $L$. The induced EMF is given by: $\mathcal{E} = BLv$ Substituting the value of $L$: $\mathcal{E} = B(\sqrt{3}r)v = \sqrt{3}Brv$

3. Equivalent resistance of the circuit ($R_{eq}$): The rod divides the ring into two segments. Let the total resistance of the ring be $R$. Consider the triangle formed by the center of the ring (O) and the two points where the rod touches the ring. Let $\theta$ be the angle between the radius to one end of the rod and the perpendicular from the center to the rod. From the geometry, $\cos \theta = \frac{\text{distance from center to rod}}{\text{radius}} = \frac{r/2}{r} = \frac{1}{2}$. Therefore, $\theta = \frac{\pi}{3}$ radians (or 60 degrees). The total angle subtended by the chord (rod) at the center is $2\theta = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$. The lengths of the two arc segments of the ring are proportional to the angles they subtend: Length of the smaller arc ($l_1$) $= r(2\pi/3)$ Length of the larger arc ($l_2$) $= 2\pi r - r(2\pi/3) = r(4\pi/3)$ The resistance of each segment is proportional to its length. Since the total resistance of the ring (circumference $2\pi r$) is $R$: Resistance of the smaller arc ($R_1$) $= \frac{l_1}{2\pi r} R = \frac{r(2\pi/3)}{2\pi r} R = \frac{1}{3}R$ Resistance of the larger arc ($R_2$) $= \frac{l_2}{2\pi r} R = \frac{r(4\pi/3)}{2\pi r} R = \frac{2}{3}R$ These two resistances are in parallel across the rod (which has zero resistance). The equivalent resistance of this parallel combination is: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(R/3)(2R/3)}{R/3 + 2R/3} = \frac{2R^2/9}{3R/3} = \frac{2R^2/9}{R} = \frac{2R}{9}$

4. Current ($I$) flowing through the rod: Using Ohm's Law, the current in the circuit is: $I = \frac{\mathcal{E}}{R_{eq}} = \frac{\sqrt{3}Brv}{2R/9} = \frac{9\sqrt{3}Brv}{2R}$

5. Magnetic force ($F_m$) on the rod: The magnetic force on a current-carrying conductor of length $L$ in a magnetic field $B$ is $F_m = ILB$. $F_m = \left(\frac{9\sqrt{3}Brv}{2R}\right) (\sqrt{3}r) B$ $F_m = \frac{9 \times 3 B^2 r^2 v}{2R} = \frac{27 B^2 r^2 v}{2R}$

6. External force ($F_{ext}$): For the rod to move with constant velocity, the external force applied must balance the magnetic force (Lenz's Law). $F_{ext} = F_m = \frac{27 B^2 r^2 v}{2R}$

7. Value of N: The problem states that the magnitude of the external force is $F = \frac{NB^2r^2v}{R}$. Comparing our calculated force with the given formula: $\frac{NB^2r^2v}{R} = \frac{27 B^2 r^2 v}{2R}$ By comparing the coefficients, we find: $N = \frac{27}{2}$

The final answer is $\boxed{27/2}$.

Explanation of the solution:

1. Calculate the effective length of the rod ($L$) within the magnetic field by using the geometry of the circle and the given distance from the center.
2. Determine the induced EMF ($\mathcal{E}$) using the formula $\mathcal{E} = BLv$.
3. Calculate the equivalent resistance ($R_{eq}$) of the ring segments, which are in parallel across the rod. The resistance of each segment is proportional to its arc length.
4. Find the current ($I$) flowing through the circuit using Ohm's law, $I = \mathcal{E}/R_{eq}$.
5. Calculate the magnetic force ($F_m$) on the rod using the formula $F_m = ILB$.
6. Since the rod moves with constant velocity, the external force must be equal in magnitude to the magnetic force.
7. Compare the derived expression for the external force with the given formula to find the value of N.