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To know which is a better choice, students must weigh both options based on important factors such as fees and ratings. As per official sources, the fees to pursue BTech at Thiagarajar College of Engineering range from INR 34,400 to INR 1.6 lakh. The fee to pursue BTech at Hindusthan College of Engineering and Technology is INR 3 lakh. In terms of fees, Thiagarajar College of Engineering is a better choice.

Moreover, as per Shiksha ratings, BTechs at Thiagarajar College of Engineering and Hindusthan College of Engineering and Technology have been rated 4.5 out of 5 and 4.3 out of 5, respectively. In terms of rating, a BTech from Thiagarajar College of Engineering is a better choice.

Note: The above information is taken from various official sources. Hence, it is indicative.

4.20 Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 $\times {10}^{-3}$ V

Electrostatic field, E = 9.0 $\times {10}^5$ V/m

Let the mass of electron = m, Charge of the electron = e, Velocity of the electron = v

Then kinetic energy of the electron = eV

$\frac{1}{2}$ m $v^2$ = eV or $\frac{e}{m}$ = $\frac{v^2}{2V}$ ………….(1)

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4.19 Magnetic field strength, B = 0.15 T

Charge on the electron, e = 1.6 $\times {10}^{-19}$ C

Mass of the electron, m = 9.1 $\times {10}^{-31}kg$

Potential difference, V = 2.0 kV = 2 $\times {10}^3$ V

Thus the kinetic energy of the electron = eV = $\frac{1}{2}$ m $v^2$ , where v = velocity of electron

v = $\sqrt{\frac{2eV}{m}}$ …….(1)

Magnetic force on the electron provides the required centripetal force of the electron. Hence, electron traces a circular path of radius r

Magnetic force on the electron = Bev

Centripetal force $\frac{m{v}^2}{r}$

Hence, Bev = $\frac{m{v}^2}{r}$

r = $\frac{mv}{Be}$ ………………(2)

From equation (1) and (2), we get

r = $\frac{m}{Be}{\left[\frac{2eV}{m}\right]}^{\frac{1}{2}}$

= $\frac{9.1\mathrm{}\times {10}^{-31}}{0.15\times 1.6\mathrm{}\times {10}^{-19}}\times$ ${\left[\frac{2\times 1.6\mathrm{}\times {10}^{-19}\times 2\mathrm{}\times {10}^3}{9.1\mathrm{}\times {10}^{-31}}\right]}^{\frac{1}{2}}$

r = 1.006 $\times {10}^{-3}$ m = 1.0 mm

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

When the field makes an angle $\theta$ of 30 $°$ wit initial velocity, the initial velocity will be,

$v_1$ = v $\sin ?\theta$

$\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{m}$ equation (2), we can write the expression for the new radius as :

$r_1=\frac{m{v}_1}{Be}=\frac{mv\sin ?\theta }{Be}=$ $\frac{9.1\mathrm{}\times {10}^{-31}}{0.15\times 1.6\mathrm{}\times {10}^{-19}}\times$ ${\left[\frac{2\times 1.6\mathrm{}\times {10}^{-19}\times 2\mathrm{}\times {10}^3}{9.1\mathrm{}\times {10}^{-31}}\right]}^{\frac{1}{2}}\times \sin ?30°$

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.

4.16 Radius of the circular coil = R

Number of turns on the coil = N

Current in the coil= I

Magnetic field at a point on its axis at a distance x is given as:

B = $\frac{{\mathit{\mu }}_0\mathit{I}{\mathit{R}}^2\mathit{N}}{2({{\mathit{x}}^2+\mathit{}{\mathit{R}}^2)}^{\frac{3}{2}}}$

where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

If the magnetic field at the centre of the coil is considered, then x = 0, then

B = $\frac{{\mathit{\mu }}_0\mathit{I}{\mathit{R}}^2\mathit{N}}{2({0+\mathit{}{\mathit{R}}^2)}^{\frac{3}{2}}}$ = $\frac{{\mathit{\mu }}_0\mathit{I}\mathit{N}}{2\mathit{R}}$

This is the familiar result for magnetic field at the centre of the coil.

Radius of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both the coils = I $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{{\left\{\right(\frac{R}{2}-d)}^2+R^2\}}^{\frac{3}{2}}+{{\left\{\right(\frac{R}{2}+d)}^2+R^2\}}^{\frac{3}{2}}\right]$

Distance between both the coils = R

Let us consider point Q at a distance d from the centre.

Then one coil is at a distance of + d from point Q

Magnetic field at point Q is given as:

$B_1$ = $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$

$\mathrm{}\mathrm{A}\mathrm{l}\mathrm{s}\mathrm{o}$ , the other coil is at a distance of $\frac{R}{2}$ – d from point Q,

Magnetic field due to this coil is given as

$B_2$ = $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$

Total magnetic field, B = $B_1$ + $B_2$

= $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$ + $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}-d)}^2+R^2)}^{\frac{3}{2}}}$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{{\left\{\right(\frac{R}{2}-d)}^2+R^2\}}^{\frac{3}{2}}+{{\left\{\right(\frac{R}{2}+d)}^2+R^2\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{\{\frac{5R^2}{4}+d^2-Rd\}}^{\frac{3}{2}}+{\{\frac{5R^2}{4}+d^2+Rd\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{5R^2}{4}\right)}^{\frac{3}{2}}\left[{\{1+\frac{4}{5}\frac{d^2}{R^2}-\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}+{\{1+\frac{4}{5}\frac{d^2}{R^2}+\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}\right]$

For d $?R$ , neglecting the factor , we get

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{5R^2}{4}\right)}^{\frac{3}{2}}\left[{\{1-\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}+{\{1+\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2R^3}\times {\left(\frac{4}{5}\right)}^{\frac{3}{2}}\left[1-\frac{6d}{5R}+1+\frac{6d}{5R}\right]$

= ${\left(\frac{4}{5}\right)}^{\frac{3}{2}}\frac{{\mu }_{0}IN}{R}$