NEET Practice Questions (MCQs) with Answers & Solutions

Given, 
Frequency of alternating electric field = $\nu$
The radius of dees = R 
Mass of protons = m 
The operating magnetic field in cyclotron = B 
The kinetic energy of the proton beam = K 
We know, 
Frequency $v=\frac{eB}{2\pi m}$ 
$KE=\frac{1}{2}m{v}^2$ and radius $R=\frac{mv}{eB}$ 
Here, velocity $v=\frac{\pi R}{T/2}=\frac{2\pi R}{T}=2\pi Rv$ 
Radius $R=\frac{m\left(2\pi Rv\right)}{eB}$ 
Magnetic field $B=\frac{2\pi mv}{e}$ 
Kinetic energy 
$K=\frac{1}{2}m{\left(2\pi Rv\right)}^2=2m{\pi }^2{v}^2{R}^2$

Given, a charge of magnitude q moving in a circle of radius R with Uniform speed v m/s 

As revolving charge is equivalent to a current, 
So, 
I = q f = q × $\frac{\omega }{2\pi }$ 
Butω =$\frac{v}{R}$ 
Where R is the radius of the circle and v is the uniform speed of the charged particle. 
Therefore, I = $\frac{qv}{2\pi R}$ 
Now, magnetic moment associated with a charged particle is given by 
μ = IA = I ×πR² 
Or μ=$\frac{qv}{2\pi R}$× $\pi R^2$
= $\frac{qvR}{2}$

The magnetic field at the center of a circular coil is directly proportional to the number of turns and the current flowing through it, and inversely proportional to the radius. Since both coils are made of the same wire, the potential difference ratio should be 4 to produce the same magnetic field.

$$\begin{gathered} \mathrm{Magnetic} \mathrm{field} \mathrm{due} \mathrm{to} \mathrm{toroid}, \mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{Ni}}{2\mathrm{\pi r}} \\ \begin{array}{l}{\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0{\mathrm{N}}_1\mathrm{i}}{2{\mathrm{\pi r}}_1}\\ {\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0{\mathrm{N}}_2\mathrm{i}}{2{\mathrm{\pi r}}_2}\\ \frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}=\frac{{\mathrm{N}}_1{\mathrm{r}}_2}{{\mathrm{N}}_2{\mathrm{r}}_1}=\frac{200}{100}\times \frac{20}{40}=1\end{array} \end{gathered}$$

When a bar magnet is bent into an arc, its magnetic dipole moment changes. The new magnetic dipole moment is directly proportional to the original dipole moment and the angle subtended by the arc at the center. For an angle of 120°, the new dipole moment is (3√3/2π)M.

If we define magnetic moment like electric dipole moment

Magnetic moment ( Ampere X $m^2$ ) = Pole strength X Distance (m)

Unit of pole strength = Ampere X meter

Magnetic field lines cannot intersect, are always closed curves, and can pass through a vacuum. Therefore, option o4 which states 'All of these' is the correct answer. Magnetic field lines represent the direction of the magnetic force at a given point and must follow certain rules to accurately depict the magnetic field.