Question

Figure shows two point charge (charge $q$, mass $m$) connected through a light inextensible string of length $2L$ and move in horizontal circle about centre of mass normal to the magnetic field $B = B_0$ with speed $V = \frac{qB_0L}{m}$, where $B_0$ is constant. (Neglect electric force between the charges which are somehow being neutralised)

The string breaks when tension is $T = \frac{3q^2B_0^2L}{4m}$. Now if the magnetic field is reduced to such a value that the string just breaks, the maximum separation between the two particles during their motion is noted as $zL$. Find $z$.

Answer 1

6

Answer 2

The problem describes two point charges (charge $q$, mass $m$) connected by a light inextensible string of length $2L$. They move in a horizontal circle about their center of mass (COM) in a magnetic field $B$. The initial speed is given as $V = \frac{qB_0L}{m}$ in a field $B_0$. Electric forces are neglected. The string breaks when the tension reaches $T_{break} = \frac{3q^2B_0^2L}{4m}$. We need to find the maximum separation between the particles after the string breaks, when the magnetic field is reduced to a value $B'$ such that the string just breaks.

1. Determine the nature of the magnetic force: The figure shows the system rotating counter-clockwise. For the charge on the right, velocity is upwards. For the charge on the left, velocity is downwards. The magnetic field is into the page (represented by 'x's). Let's assume $q$ is positive. For the right charge: $\vec{v}$ is in +y, $\vec{B}$ is in -z. Magnetic force $\vec{F}_B = q(\vec{v} \times \vec{B})$ points in the -x direction (towards the COM). For the left charge: $\vec{v}$ is in -y, $\vec{B}$ is in -z. Magnetic force $\vec{F}_B = q(\vec{v} \times \vec{B})$ points in the +x direction (towards the COM). In this case, the magnetic force is inward, contributing to the centripetal force. The equation for tension $T$ would be: $T + qVB = \frac{mV^2}{L}$ If $q$ were negative, the magnetic force would be outward, and the equation would be $T - |q|VB = \frac{mV^2}{L}$.

Let's test the initial conditions provided: $V = \frac{qB_0L}{m}$ and $B = B_0$. If magnetic force is inward: $T = \frac{m}{L} \left(\frac{qB_0L}{m}\right)^2 - q\left(\frac{qB_0L}{m}\right)B_0 = \frac{q^2B_0^2L}{m} - \frac{q^2B_0^2L}{m} = 0$. If magnetic force is outward: $T = \frac{m}{L} \left(\frac{qB_0L}{m}\right)^2 + q\left(\frac{qB_0L}{m}\right)B_0 = \frac{q^2B_0^2L}{m} + \frac{q^2B_0^2L}{m} = \frac{2q^2B_0^2L}{m}$. The breaking tension is $T_{break} = \frac{3q^2B_0^2L}{4m}$. If the magnetic force is inward, $T=0$, so the string would never break. If the magnetic force is outward, $T = \frac{2q^2B_0^2L}{m}$, which is greater than $T_{break}$. This means the string would have already broken in the initial setup.

This implies that the initial speed $V = \frac{qB_0L}{m}$ is just a reference value, and the question is about a new scenario where the string just breaks. The problem implies that the magnetic force must be outward for the string to break at a specific tension. So, we assume the magnetic force is outward, meaning $q$ is negative, or the direction of $B$ or rotation is opposite to what a positive $q$ would imply for an inward force. We will use $|q|$ for the magnitude of charge and assume the force is $F_B = |q|VB$ acting outwards.

2. Condition for string breaking: When the string just breaks, the tension is $T_{break}$, the magnetic field is $B'$, and the speed of each particle is $V'$. The radius of rotation is still $L$. $T_{break} - |q|V'B' = \frac{mV'^2}{L}$ Substitute $T_{break} = \frac{3q^2B_0^2L}{4m}$: $\frac{3q^2B_0^2L}{4m} - |q|V'B' = \frac{mV'^2}{L}$ (Equation 1)

3. Motion after string breaks: When the string breaks, the particles are at a distance $2L$ from each other. Each particle has a speed $V'$ and is at a distance $L$ from the original COM. Since the string breaks, the tension becomes zero. Each particle now moves under the influence of the magnetic force only. The magnetic force $F_B = |q|V'B'$ is outward (away from the original COM). Since the magnetic force is perpendicular to the velocity, it provides the centripetal force for each particle to move in its own circular path. Let $R_c$ be the radius of the circular path of each particle. $|q|V'B' = \frac{mV'^2}{R_c}$ $R_c = \frac{mV'}{|q|B'}$ (Equation 2)

4. Determine $V'$ and $B'$ from the breaking condition: The problem states "if the magnetic field is reduced to such a value that the string just breaks". This implies that the initial speed $V = \frac{qB_0L}{m}$ is maintained, or that this relationship between speed and field is maintained. The most reasonable interpretation is that the initial speed $V = \frac{qB_0L}{m}$ is the speed at which the system was moving, and the magnetic field $B_0$ is the initial field. Now, the field is reduced to $B'$, and the speed is adjusted to $V'$ such that the string just breaks. However, if the speed $V'$ is determined by the magnetic field $B'$ (as in $V' = \frac{|q|B'L}{m}$), then substituting this into Equation 1: $\frac{3q^2B_0^2L}{4m} - |q|\left(\frac{|q|B'L}{m}\right)B' = \frac{m}{L}\left(\frac{|q|B'L}{m}\right)^2$ $\frac{3q^2B_0^2L}{4m} - \frac{q^2B'^2L}{m} = \frac{q^2B'^2L}{m}$ $\frac{3q^2B_0^2L}{4m} = \frac{2q^2B'^2L}{m}$ $\frac{3}{4}B_0^2 = 2B'^2$ $B'^2 = \frac{3}{8}B_0^2 \implies B' = \frac{\sqrt{3}}{2\sqrt{2}}B_0 = \frac{\sqrt{6}}{4}B_0$. And $V' = \frac{|q|B'L}{m} = \frac{|q|}{m}\left(\frac{\sqrt{6}}{4}B_0\right)L = \frac{\sqrt{6}}{4}\frac{|q|B_0L}{m}$.

5. Calculate the radius of curvature $R_c$ after breaking: Substitute $V'$ and $B'$ into Equation 2: $R_c = \frac{mV'}{|q|B'} = \frac{m}{|q|B'} \left(\frac{|q|B'L}{m}\right) = L$. This result implies that after the string breaks, each particle continues to move in a circle of radius $L$.

6. Maximum separation: Initially, the particles are at a distance $2L$. Each particle moves in a circle of radius $L$ about the COM. When the string breaks, each particle starts moving in its own circular path of radius $R_c = L$. The initial positions of the particles (say, $P_1$ and $P_2$) are $L$ apart from the COM. Let the COM be at the origin $(0,0)$. $P_1$ is at $(-L, 0)$ and $P_2$ is at $(L, 0)$. At the moment of breaking, $P_1$ has velocity in -y direction and $P_2$ has velocity in +y direction. The magnetic force on $P_1$ is in -x direction (outward from original COM). So $P_1$ moves in a circle centered at $(-2L, 0)$ with radius $L$. Its path is $(x+2L)^2 + y^2 = L^2$. The magnetic force on $P_2$ is in +x direction (outward from original COM). So $P_2$ moves in a circle centered at $(2L, 0)$ with radius $L$. Its path is $(x-2L)^2 + y^2 = L^2$.

Let's re-examine the center of the circular path. For $P_1$ at $(-L, 0)$ with velocity $(0, -V')$, the magnetic force is $(|q|V'B', 0)$. The center of the circle for $P_1$ is $C_1 = (-L - R_c, 0) = (-L-L, 0) = (-2L, 0)$. For $P_2$ at $(L, 0)$ with velocity $(0, V')$, the magnetic force is $(-|q|V'B', 0)$. The center of the circle for $P_2$ is $C_2 = (L + R_c, 0) = (L+L, 0) = (2L, 0)$.

The path of $P_1$ is a circle centered at $(-2L, 0)$ with radius $L$. The path of $P_2$ is a circle centered at $(2L, 0)$ with radius $L$. The particles always move on these circles. The minimum separation between the particles is when they are at their closest points. $P_1$ at $(-L, 0)$ and $P_2$ at $(L, 0)$. Separation is $2L$. The maximum separation occurs when they are at their farthest points. For $P_1$, the x-coordinates range from $-2L-L = -3L$ to $-2L+L = -L$. For $P_2$, the x-coordinates range from $2L-L = L$ to $2L+L = 3L$. The maximum separation occurs when $P_1$ is at $(-3L, 0)$ and $P_2$ is at $(3L, 0)$. The maximum separation is $3L - (-3L) = 6L$.

Comparing this with $zL$, we get $z=6$.