Question

a sphere of radius r is filled with an electric field which points everywhere in the z direction, which defines the axis of the sphere. while the direction is constant, the magnitude of the field is zero on the axis and increases linearly with distance from the axis. when the magnitude of the electric field changes rapidly in time, a magnetic field, b₁, is measured at a point outside of the sphere which lies in the equatorial plane of the sphere (perpendicular to the z - axis) at a distance d from the center of the sphere. if the sphere is replaced by a new sphere which has twice the volume of the original, but nothing else is changed, how will the magnetic field, b₂, at the same location, still a distance d from the center, compare to the strength of the original magnetic field? the location at distance d still lies outside the new sphere and the new sphere is still filled by an electric field which varies in time and space in the same way as the original field. b₂ = b₁ b₂ = 8 b₁ b₂ = 4 b₁ b₂ = 2 b₁

Explanation:

Step1: Express the electric field

Let the electric field be \(E = kr\) (where \(r\) is the distance from the \(z -\)axis and \(k\) is a constant). Using cylindrical coordinates, for a sphere of radius \(R\), we can calculate the displacement current. The displacement current density is \(j_d=\epsilon_0\frac{\partial E}{\partial t}\).

Step2: Calculate the displacement current for the original sphere

The displacement current \(I_{d1}\) through the original sphere:
We use the formula \(I_d=\epsilon_0\int\frac{\partial E}{\partial t}dA\). In cylindrical coordinates, \(dA = 2\pi rdr\) (for a thin cylindrical shell of radius \(r\) and thickness \(dr\) in the cross - section of the sphere).
\(I_{d1}=\epsilon_0\frac{\partial k}{\partial t}\int_{0}^{R}r\cdot2\pi rdr=\epsilon_0\frac{\partial k}{\partial t}\cdot2\pi\int_{0}^{R}r^{2}dr=\epsilon_0\frac{\partial k}{\partial t}\cdot2\pi\frac{R^{3}}{3}\)

Step3: Calculate the radius of the new sphere

The volume of a sphere \(V = \frac{4}{3}\pi R^{3}\). If \(V_2 = 2V_1\), then \(\frac{4}{3}\pi R_{2}^{3}=2\times\frac{4}{3}\pi R_{1}^{3}\), so \(R_{2}=2^{\frac{1}{3}}R_{1}\)

Step4: Calculate the displacement current for the new sphere

\(I_{d2}=\epsilon_0\frac{\partial k}{\partial t}\cdot2\pi\frac{R_{2}^{3}}{3}\). Substituting \(R_{2}=2^{\frac{1}{3}}R_{1}\) into the formula for \(I_{d2}\), we get \(I_{d2}=\epsilon_0\frac{\partial k}{\partial t}\cdot2\pi\frac{(2^{\frac{1}{3}}R_{1})^{3}}{3}= 2\epsilon_0\frac{\partial k}{\partial t}\cdot2\pi\frac{R_{1}^{3}}{3}=2I_{d1}\)

Step5: Use Ampere - Maxwell law

Ampere - Maxwell law is \(\oint\vec{B}\cdot d\vec{l}=\mu_0I_{enc}\). For a point outside the sphere (\(d > R\)), \(\oint\vec{B}\cdot d\vec{l}=B(2\pi d)\) and \(I_{enc}=I_d\) (displacement current).
For the original sphere: \(B_1(2\pi d)=\mu_0I_{d1}\), so \(B_1=\frac{\mu_0I_{d1}}{2\pi d}\)
For the new sphere: \(B_2(2\pi d)=\mu_0I_{d2}\). Since \(I_{d2} = 2I_{d1}\), then \(B_2=\frac{\mu_0I_{d2}}{2\pi d}=\frac{\mu_0(2I_{d1})}{2\pi d}=2B_1\)

Answer:

\(B_2 = 2B_1\)