Reasoning: Magnetic induction or magnetic flux density $\mathbf{B}$ is measured in tesla $\mathbf{(T)}$ in the SI system.

Reasoning: $\mathbf{F = BIL}\mathbf{\ }\mathbf{sin\theta}$
Given: $\mathbf{B = 2.0}\mathbf{\ }\mathbf{T}$, $\mathbf{I = 3.2}\mathbf{\ }\mathbf{A}$, $\mathbf{L = 0.5}\mathbf{\ }\mathbf{m}$, $\mathbf{\theta =}\mathbf{90}^{\mathbf{o}}$
$\mathbf{F = (2.0)(3.2)(0.5)}\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{90}^{\mathbf{o}}$ $\mathbf{\Longrightarrow}$ $\mathbf{F = 3.2}\mathbf{\ }\mathbf{N}$

Reasoning: The magnetic force on both particles has the same magnitude:
$\mathbf{F =}$ $\mathbf{qvB}$ $\mathbf{\Longrightarrow}$ Since, $\mathbf{a =}$ $\frac{\mathbf{F}}{\mathbf{m}}$ $\mathbf{\Longrightarrow}$ $\frac{\mathbf{a}_{\mathbf{e}}}{\mathbf{a}_{\mathbf{p}}}\mathbf{=}\frac{\mathbf{m}_{\mathbf{p}}}{\mathbf{m}_{\mathbf{e}}}$
Therefore, the electron has much greater acceleration because its mass is smaller.

Reasoning: The velocity has two components:
● A perpendicular component that produces circular motion. ● A parallel component that moves the particle along the magnetic field. The combined motion produces a helical path.

Reasoning: The radius of the circular path is $\mathbf{r =}$ $\frac{\mathbf{mv}}{\mathbf{qB}}\mathbf{\ }$.
Thus, $\mathbf{r \propto}$ $\frac{\mathbf{1}}{\mathbf{B}}$ When the magnetic field is halved. $\mathbf{\Longrightarrow}$ $\mathbf{r}^{\mathbf{'}}\mathbf{=}$ $\frac{\mathbf{mv}}{\mathbf{q(B/2)}}\mathbf{= 2r}$

Reasoning: A magnetic field exerts force on a moving charged particle:
$\mathbf{F =}$ $\mathbf{qvB}\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{\theta}$
A neutron has no electric charge, so: $\mathbf{q = 0}$ $\mathbf{\Longrightarrow}$ $\mathbf{F = 0}$ Therefore, it is not deflected.

Reasoning: The radius is $\mathbf{r =}$ $\frac{\mathbf{mv}}{\mathbf{qB}}$.
For a proton $\mathbf{\Longrightarrow}$ $\mathbf{r}_{\mathbf{p}}\mathbf{=}$ $\frac{\mathbf{m}_{\mathbf{p}}\mathbf{v}}{\mathbf{eB}}$ For an alpha particle $\mathbf{\Longrightarrow}$ $\mathbf{m}_{\mathbf{\alpha}}\mathbf{= 4}\mathbf{m}_{\mathbf{p}}\mathbf{\ }$, $\mathbf{q}_{\mathbf{\alpha}}\mathbf{= 2e}$
$\mathbf{r}_{\mathbf{\alpha}}\mathbf{=}$ $\frac{\mathbf{4}\mathbf{m}_{\mathbf{p}}\mathbf{v}}{\mathbf{2eB}}\mathbf{=}\frac{\mathbf{2}\mathbf{m}_{\mathbf{p}}\mathbf{v}}{\mathbf{eB}}$ $\mathbf{\Longrightarrow}$ Therefore, $\frac{\mathbf{r}_{\mathbf{p}}}{\mathbf{r}_{\mathbf{\alpha}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Reasoning: For a charged particle moving perpendicular to a magnetic field:
$\mathbf{F}_{\mathbf{B}}\mathbf{= qvB}$ $\mathbf{\ldots\ldots\ldots}\mathbf{(i)}$
The centripetal force required for circular motion is:
$\mathbf{F}_{\mathbf{c}}\mathbf{=}$ $\frac{\mathbf{m}\mathbf{v}^{\mathbf{2}}}{\mathbf{R}}$ $\mathbf{\ldots\ldots\ldots}\mathbf{(i}\mathbf{i}\mathbf{)}$
Equating $\mathbf{(i)}$ and $\mathbf{(ii)}$ equations. $\mathbf{qvB =}$ $\frac{\mathbf{m}\mathbf{v}^{\mathbf{2}}}{\mathbf{R}}$ $\mathbf{\Longrightarrow}$ $\mathbf{qB =}$ $\frac{\mathbf{mv}}{\mathbf{R}}$
Therefore, $\mathbf{R =}$ $\frac{\mathbf{mv}}{\mathbf{qB}}$
The kinetic energy of the particle is $\mathbf{K =}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}\mathbf{v}^{\mathbf{2}}$.
Multiply both sides by $\mathbf{2}$. $\mathbf{\Longrightarrow}$ $\mathbf{2K = m}\mathbf{v}^{\mathbf{2}}$
Divide by $\mathbf{v}^{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{2K}}{\mathbf{m}}$. $\mathbf{\Longrightarrow}$ Taking square root: $\mathbf{v =}$ $\sqrt{\frac{\mathbf{2K}}{\mathbf{m}}}$
We have $\mathbf{R =}$ $\frac{\mathbf{mv}}{\mathbf{qB}}$ $\mathbf{\Longrightarrow}$ Substitute: $\mathbf{v}\mathbf{\ }\mathbf{=}$ $\sqrt{\frac{\mathbf{2K}}{\mathbf{m}}}$
So, $\mathbf{R =}$ $\frac{\mathbf{m}}{\mathbf{qB}}\mathbf{\ }\sqrt{\frac{\mathbf{2K}}{\mathbf{m}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R =}$ $\frac{\sqrt{\mathbf{2mK}}}{\mathbf{qB}}$ $\mathbf{\Longrightarrow}$ Therefore, $\mathbf{R \propto}\sqrt{\mathbf{K}}$
When kinetic energy is doubled: $\mathbf{K}^{\mathbf{'}}\mathbf{= 2K}$
The new radius is $\mathbf{R}^{\mathbf{'}}\mathbf{=}$ $\frac{\sqrt{\mathbf{2m}\mathbf{K}^{\mathbf{'}}}}{\mathbf{qB}}$.
Substitute $\mathbf{K}^{\mathbf{'}}\mathbf{= 2K}$. $\mathbf{\Longrightarrow}$ $\mathbf{R}^{\mathbf{'}}\mathbf{=}$ $\frac{\sqrt{\mathbf{2m}\mathbf{\ }\mathbf{(2K)}}}{\mathbf{qB}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R}^{\mathbf{'}}\mathbf{=}$ $\frac{\sqrt{\mathbf{4mK}}}{\mathbf{qB}}$
To compare it with the original radius.
$\frac{\mathbf{R}^{\mathbf{'}}}{\mathbf{R}}\mathbf{=}\sqrt{\frac{\mathbf{K}^{\mathbf{'}}}{\mathbf{K}}}$ $\mathbf{\Longrightarrow}$ $\frac{\mathbf{R}^{\mathbf{'}}}{\mathbf{R}}\mathbf{=}\sqrt{\frac{\mathbf{2K}}{\mathbf{K}}}$ $\mathbf{\Longrightarrow}$ $\frac{\mathbf{R}^{\mathbf{'}}}{\mathbf{R}}\mathbf{=}\sqrt{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ Hence, $\mathbf{R}^{\mathbf{'}}\mathbf{=}\sqrt{\mathbf{2}}\mathbf{\ }\mathbf{R}$

Reasoning: For motion perpendicular to the magnetic field:
$\mathbf{F = qvB}$ $\mathbf{\Longrightarrow}$ $\mathbf{q =}$ $\frac{\mathbf{F}}{\mathbf{vB}}$ $\mathbf{\Longrightarrow}$ $\mathbf{q =}$ $\frac{\mathbf{8.1}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{10}^{\mathbf{- 16}}}{\mathbf{(}\mathbf{3.1}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{10}^{\mathbf{5}}\mathbf{)}\mathbf{\ }\mathbf{(}\mathbf{5.4}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{10}^{\mathbf{- 5}}\mathbf{)}}$
$\mathbf{q =}$ $\frac{\mathbf{8.1}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{10}^{\mathbf{- 16}}}{\mathbf{16.74}}$ $\mathbf{\Longrightarrow}$ $\mathbf{q \approx 4.8 \times}\mathbf{10}^{\mathbf{- 17}}\mathbf{C}$

Reasoning: The proton moves along the axis of the solenoid, so its velocity is parallel or antiparallel to the magnetic field.
$\mathbf{F =}$ $\mathbf{qvB}\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{\theta}$
Here, $\mathbf{\theta =}\mathbf{0}^{\mathbf{o}}$ or $\mathbf{180}^{\mathbf{o}}$. $\mathbf{\Longrightarrow}$ $\mathbf{F = qvB}\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{0}^{\mathbf{o}}\mathbf{= 0}$ Therefore, no magnetic force acts on the proton.

Reasoning: Magnetic flux is $\mathbf{\Phi = BA}\mathbf{\ }\mathbf{\cos}\mathbf{\ }\mathbf{\theta}$.
Here, $\mathbf{\theta}$ is the angle between the magnetic field and the normal to the area. Flux is maximum when $\mathbf{\theta =}\mathbf{0}^{\mathbf{o}}$. Therefore, the magnetic field must be perpendicular to the plane of the area.

Reasoning: According to Faraday’s law: $\mathcal{E = -}\mathbf{\ }\mathbf{N}$ $\frac{\mathbf{\Delta\Phi}}{\mathbf{\Delta t}}$
An e.m.f. is induced whenever the magnetic flux through a coil changes, whether it increases or decreases.

Reasoning: According to Lenz’s law, the induced current always opposes the change that produces it. This prevents energy from being created without external work and therefore agrees with the law of conservation of energy.

Reasoning: When the north pole of the magnet moves toward the metallic ring, the magnetic flux through the ring increases. According to Lenz’s law, an induced current flows in the ring to oppose this increase. Therefore, the side of the ring facing the magnet becomes a north pole. Since like poles repel, the ring moves away from the magnet.

Reasoning: For a long hollow conducting pipe carrying direct current, the current flows through the conducting material. Inside the hollow region, an Amperian loop encloses no current, so the magnetic field there is zero. Outside the pipe, the loop encloses the total current, so a magnetic field exists around the pipe.
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