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Question: To find the force on the electric dipole, we first need to determine the induced electric field `E` ...

To find the force on the electric dipole, we first need to determine the induced electric field E inside the cylinder due to the changing magnetic flux.

1. Induced Electric Field (E):

According to Faraday's Law of Induction, the induced EMF ε\varepsilon around a closed loop is related to the rate of change of magnetic flux Φ\Phi through the loop by:

ε=−dΦdt\varepsilon = - \frac{d\Phi}{dt}

The magnetic flux Φ\Phi through a circular area of radius R (the cross-section of the cylinder) is Φ=B⋅A=B⋅πR2\Phi = B \cdot A = B \cdot \pi R^2.

So, the induced EMF on the circumference of the cylinder is:

ε=−ddt(BπR2)=−πR2dBdt\varepsilon = - \frac{d}{dt}(B \pi R^2) = - \pi R^2 \frac{dB}{dt}

From this, we can express dB/dt:

dBdt=−επR2(1)\frac{dB}{dt} = - \frac{\varepsilon}{\pi R^2} \quad (1)

The induced electric field E at a distance r from the axis of the cylinder can be found using the integral form of Faraday's Law:

∮E⃗⋅dl⃗=−dΦrdt\oint \vec{E} \cdot d\vec{l} = - \frac{d\Phi_r}{dt}

For a circular path of radius r inside the cylinder, the induced electric field E will be tangential (azimuthal) due to symmetry.

E⋅(2πr)=−ddt(Bπr2)E \cdot (2\pi r) = - \frac{d}{dt}(B \pi r^2) E⋅(2πr)=−πr2dBdtE \cdot (2\pi r) = - \pi r^2 \frac{dB}{dt}

Solving for E:

E=−r2dBdtE = - \frac{r}{2} \frac{dB}{dt}

The direction of E is azimuthal. Let's assume the cylinder's axis is the z-axis. The electric field vector at a point (x, y) in the xy-plane is given by:

E⃗=12dBdt(yi^−xj^)\vec{E} = \frac{1}{2} \frac{dB}{dt} (y \hat{i} - x \hat{j})

(This corresponds to a clockwise field if dB/dt > 0).

2. Force on the Electric Dipole:

An electric dipole consists of charges -q and +q separated by a distance l. The dipole moment is $\vec{p} = q\vec{l}$, directed from -q to +q.

Let's assume the dipole is centered at the origin of the cylinder and oriented along the x-axis, as suggested by the diagram. So, -q is at (-l/2, 0) and +q is at (l/2, 0). The dipole moment is $\vec{p} = ql \hat{i}$.

The force on an electric dipole in a non-uniform electric field $\vec{E}$ is given by:

F⃗=(p⃗⋅∇)E⃗\vec{F} = (\vec{p} \cdot \nabla) \vec{E}

Here, $\vec{p} = ql \hat{i}$ and $\nabla = \frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}$.

So, $\vec{p} \cdot \nabla = (ql \hat{i}) \cdot (\frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}) = ql \frac{\partial}{\partial x}$.

Now, apply this operator to $\vec{E}$:

F⃗=ql∂∂x[12dBdt(yi^−xj^)]\vec{F} = ql \frac{\partial}{\partial x} \left[ \frac{1}{2} \frac{dB}{dt} (y \hat{i} - x \hat{j}) \right]

Since y and dB/dt are not functions of x:

F⃗=ql12dBdt∂∂x(yi^−xj^)\vec{F} = ql \frac{1}{2} \frac{dB}{dt} \frac{\partial}{\partial x} (y \hat{i} - x \hat{j}) F⃗=ql12dBdt(0i^−1j^)\vec{F} = ql \frac{1}{2} \frac{dB}{dt} (0 \hat{i} - 1 \hat{j}) F⃗=−ql2dBdtj^(2)\vec{F} = - \frac{ql}{2} \frac{dB}{dt} \hat{j} \quad (2)

3. Substitute dB/dt from (1) into (2):

Substitute $\frac{dB}{dt} = - \frac{\varepsilon}{\pi R^2}$ into the force equation:

F⃗=−ql2(−επR2)j^\vec{F} = - \frac{ql}{2} \left( - \frac{\varepsilon}{\pi R^2} \right) \hat{j} F⃗=qlε2πR2j^\vec{F} = \frac{ql\varepsilon}{2\pi R^2} \hat{j}

The magnitude of the force on the dipole is $\frac{ql\varepsilon}{2\pi R^2}$. The direction of the force is perpendicular to the dipole axis (y-direction if dipole is along x-axis).

Answer

qlε2πR2\frac{ql\varepsilon}{2\pi R^2}

Explanation

Solution

  1. Relate EMF to changing magnetic field: Faraday's Law ε=−dΦ/dt=−πR2(dB/dt)\varepsilon = -d\Phi/dt = -\pi R^2 (dB/dt) is used to find (dB/dt)(dB/dt) in terms of ε\varepsilon and cylinder radius R.

  2. Determine induced electric field: The induced electric field E⃗\vec{E} at a distance r from the axis is found using ∮E⃗⋅dl⃗=−dΦr/dt\oint \vec{E} \cdot d\vec{l} = -d\Phi_r/dt, yielding E⃗=12dBdt(yi^−xj^)\vec{E} = \frac{1}{2} \frac{dB}{dt} (y \hat{i} - x \hat{j}).

  3. Calculate force on dipole: The force on an electric dipole p⃗\vec{p} in a non-uniform electric field E⃗\vec{E} is F⃗=(p⃗⋅∇)E⃗\vec{F} = (\vec{p} \cdot \nabla) \vec{E}. Assuming the dipole p⃗=qli^\vec{p} = ql \hat{i} is centered at the origin, the force is calculated by applying the operator (ql∂∂x)(ql \frac{\partial}{\partial x}) to E⃗\vec{E}.

  4. Substitute and simplify: Substitute the expression for (dB/dt)(dB/dt) into the force equation to get the final force in terms of q, l, ε\varepsilon, and R.