Subfield: Electromagnetism

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: A uniform spherical metallic ball of mass $m$, resistivity $\rho$, and radius $R$ is kept on a smooth friction-less horizontal ground. A horizontal uniform, constant magnetic field $B$ exists in the space parallel to the surface of ground. The ball was suddenly given an impulse perpendicular to magnetic field such that ball begin to move with velocity $v$ without losing the contact with ground. Find the time in seconds required to reduce its velocity by half.
Numerical Quantities: $m=2 \mathrm{~kg}, 4 \pi \epsilon_{0} R^{3} B^{2}=3 \mathrm{~kg}, \rho=10^{9} \Omega \mathrm{m}, v=\pi \mathrm{m} / \mathrm{s}$.

Solution:
WLOG assuming magnetic field to be into the plane (negative z-axis) and velocity of the block along x-axis. Let any time $\mathrm{t}$ ball has velocity $\mathrm{v}$ and surface charge $\sigma \cos \theta$, where $\theta$ is measured from $\mathrm{Y}$-axis. As the ball is moving it will also generate and electric field $\mathrm{E}=\mathrm{vB}$ along positive Y-axis. Which will be opposed by the electric field of ball. Also we know the electric filed generated by the ball is $\frac{\sigma}{3 \epsilon_{0}}$ in negative $\mathrm{Z}$-axis. As this charge distribution will arise form a vertical electric field and the subsequent induced charges will also produce only a vertical electric field only thus our assumption about charge distribution and net electric field must be true.
Now at any point on the surface of the sphere rate of increase in surface charge density is given by-



$$
\begin{gathered}
J d A \cos \theta=\frac{d(\sigma \cos \theta)}{d t} d A \\
J=\frac{d \sigma}{d t} \\
E-\frac{\sigma}{3 \epsilon_{0}}=\rho \frac{d \sigma}{d t} \\
v B-\frac{\sigma}{3 \epsilon_{0}}=\rho \frac{d \sigma}{d t}
\end{gathered}
\tag{44}
$$

As the magnetic field is uniform to calculate force path of the current will not matter. Hence assuming it to be straight line between two points located at $\theta$ and $-\theta$. So the force can be written as -

$$
\begin{gathered}
d F=B \times d I \times l \\
d F=B(2 \pi(R \sin \theta) \times R d \theta \times J)(2 R \cos (\theta)) \\
F=4 \pi R^{3} B \frac{d \sigma}{d t} \int_{0}^{\pi / 2} \sin \theta \cos ^{2} \theta d \theta \\
F=\frac{4}{3} \pi R^{3} B \frac{d \sigma}{d t}
\end{gathered}
\tag{45}
$$

Now writing force equation on the sphere we have

$$
\begin{gathered}
F=-m \frac{d v}{d t} \\
\frac{4}{3} \pi R^{3} B \frac{d \sigma}{d t}=-m \frac{d v}{d t} \\
\frac{4}{3} \pi R^{3} B \int_{0}^{\sigma} d \sigma=-m \int_{v_{0}}^{v} d v \\
\frac{4}{3} \pi R^{3} B \sigma=m v_{0}-m v
\end{gathered}
\tag{46}
$$

Solving equation 1 and 3 gives us

$$
\left(m+4 \pi R^{3} B^{2} \epsilon_{0}\right) v-m v_{0}=-3 m \rho \epsilon_{0} \frac{d v}{d t}
$$

Integrating it from $v_{0}$ to $\frac{v_{0}}{2}$ gives

$$

t=\frac{3 m \rho \epsilon_{0}}{\left(m+4 \pi R^{3} B^{2} \epsilon_{0}\right)} \ln \left(\frac{8 \pi R^{3} B^{2} \epsilon_{0}}{4 \pi R^{3} B^{2} \epsilon_{0}-m}\right)
\tag{47}
$$
$$
t=\frac{6 \ln (6)}{5} \rho \epsilon_{0}=0.019 \mathrm{~s}
\tag{48}
$$

Final answer: $0.019$ s