Correct answer option is (C)

To solve for the ratio of the radii of deutron and proton paths in a magnetic field, we use the formula for the radius $r$ of the circular path of a charged particle moving perpendicular to a uniform magnetic field:

$r=\frac{mv}{qB}$

where:

• $m$ is the mass of the particle,
• $v$ is the velocity of the particle,
• $q$ is the charge of the particle, and
• $B$ is the magnetic field strength.

The proton and the deutron are given to have the same kinetic energy. The kinetic energy $K$ of a particle is given by:

$K=\frac{1}{2}m{v}^2$

From the kinetic energy, we can express the velocity as:

$v=\sqrt{\frac{2K}{m}}$

Since both particles have the same kinetic energy and the charge of the deutron is the same as the charge of the proton $q=e$, but the mass of the deutron is twice that of the proton ($m_d=2m_p$), substituting the expression for $v$ in the radius formula, we get:

For the deutron:

$r_d=\frac{m_d\sqrt{2K/m_d}}{eB}=\sqrt{\frac{2K}{e{B}^2}}\cdot \sqrt{m_d}$

For the proton ($m_p=m$):

$r_p=\sqrt{\frac{2K}{e{B}^2}}\cdot \sqrt{m_p}$

The ratio of the radius of the deutron path $r_d$ to the radius of the proton path $r_p$ is therefore:

$\frac{r_d}{r_p}=\frac{\sqrt{m_d}}{\sqrt{m_p}}=\sqrt{\frac{2m_p}{m_p}}=\sqrt{2}$

So, the correct answer is:

Option C: $\sqrt{2}:1$.

Correct answer option is (C)

The correct expressions for the electrostatic force, ${\overrightarrow{F}}_1$, and the magnetic force, ${\overrightarrow{F}}_2$, acting on a charge $q$ moving with velocity $\overrightarrow{v}$, are given by the Lorentz force law. This law states that the total force acting on a charged particle in both an electric field and a magnetic field is the sum of an electrostatic force due to the electric field and a magnetic force due to the magnetic field.

The electrostatic force is given by:

${\overrightarrow{F}}_1=q\overrightarrow{E}$

where $\overrightarrow{E}$ is the electric field.

The magnetic force is given by:

${\overrightarrow{F}}_2=q(\overrightarrow{v}\times \overrightarrow{B})$

where $\overrightarrow{B}$ is the magnetic field, and $\times$ denotes the cross product, indicating that the magnetic force is perpendicular both to the direction of the velocity of the charge and the direction of the magnetic field.

Therefore, the correct option is:

Option C: ${\overrightarrow{F}}_1=q\overrightarrow{E},{\overrightarrow{F}}_2=q(\overrightarrow{V}\times \overrightarrow{B})$

Options A, B, and D are incorrect because they do not accurately reflect the definitions of electrostatic and magnetic forces as described by the Lorentz force law.

Correct answer option is (D)

For this scenario, it's important to recall how magnetic fields influence the motion of charged particles, and the configuration of the magnetic field within a solenoid. Inside a long solenoid, the magnetic field lines run parallel to the axis of the solenoid. The strength of this field is uniform and depends on the current running through the solenoid's coils and the number of turns per unit length but is independent of the position inside the solenoid as long as one is sufficiently far from the ends.

The force experienced by a charged particle moving in a magnetic field is given by the Lorentz force equation: $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$ where

• $q$ is the charge of the particle,


• $\overrightarrow{v}$ is the velocity of the particle, and


• $\overrightarrow{B}$ is the magnetic field.

For an electron moving along the axis inside a solenoid:

• The magnetic field ($\overrightarrow{B}$) inside the solenoid is parallel to the axis of the solenoid.


• The velocity of the electron ($\overrightarrow{v}$) is also parallel to the axis of the solenoid and hence parallel to $\overrightarrow{B}$.

Since the cross product of two parallel vectors (in this case, $\overrightarrow{v}$ and $\overrightarrow{B}$) is zero, the Lorentz force ($\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$) acting on the electron will be zero. Therefore, the electron will not experience any force due to the magnetic field of the solenoid, as there is no component of its velocity that is perpendicular to the magnetic field within the solenoid.

Given the above explanation, the correct option is:

Option D: the electron will continue to move with uniform velocity along the axis of the solenoid.

This is because, in the absence of any force acting on it, the electron will maintain its state of motion according to Newton's first law of motion, which states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

Correct answer option is (D)

$\begin{array}{rl}& \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}}=\frac{\sqrt{2\mathrm{m}(\mathrm{KE})}}{\mathrm{qB}}=\frac{\sqrt{2\mathrm{mqV}}}{\mathrm{qB}}\\ & \mathrm{R}\propto \sqrt{\mathrm{m}}\\ & \mathrm{m}\propto {\mathrm{R}}^2\\ & \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}={\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right)}^2\end{array}$

Correct answer option is (C)

Net force on particle must be zero i.e. $\mathrm{q}\overrightarrow{\mathrm{E}}+\mathrm{q}\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}=0$

Possible cases are

(i) $\overrightarrow{\mathrm{E}}\mathrm{\&}\overrightarrow{\mathrm{B}}=0$

(ii) $\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}=0,\overrightarrow{\mathrm{E}}=0$

(iii) $q\overrightarrow{\mathrm{E}}=-\mathrm{q}\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}$

$\overrightarrow{\mathrm{E}}\ne 0\mathrm{\&}\overrightarrow{\mathrm{B}}\ne 0$