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A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic field $\vec{B}$ . The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $\vec{B}$ is :

25 kev

50 kev

200 kev

100 kev

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Explanation

(d)

$r = \frac{\sqrt{2 mk}}{qB} \Rightarrow K \propto \frac{q^{2}}{m} \Rightarrow \frac{k_{p}}{k_{d}} = {(\frac{q_{p}}{q_{d}})}^{2} \times \frac{m_{d}}{m_{p}} = ({\frac{1}{1}}^{2}) \times \frac{2}{1} = \frac{2}{1} \Rightarrow k_{p} = 2 \times 50 = 100 k e v$

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Two thin long parallel wires separated by a distance b are carrying a current i amp each. The magnitude of the force per unit length exerted by one wire on the other is

$\frac{μ_{0} i^{2}}{b^{2}}$

$\frac{μ_{0} i^{2}}{2 πb}$

$\frac{μ_{0} i}{2 πb}$

$\frac{μ_{0} i}{2 {πb}^{2}}$

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Explanation

(b) Force per unit length $= \frac{μ_{0}}{4 π} . \frac{2 i_{1} i_{2}}{r} = \frac{μ_{0}}{2 π} . \frac{i^{2}}{b}$

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To make the field radial in a moving coil galvanometer :

The number of turns in the coil is increased

Magnet is taken in the form of horse-shoe

Poles are cylindrically cut

The coil is wounded on the aluminum frame

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Explanation

(c)

Poles are cylindrically cut

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In a moving coil galvanometer, the deflection of the coil $θ$ is related to the electrical current i by the relation

$i \propto \tan θ$

$i \propto θ$

$i \propto θ^{2}$

$i \propto \sqrt{θ}$

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Explanation

(b) $i = \frac{C θ}{N A B} \Rightarrow i \propto θ$

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A moving coil galvanometer has N number of turns in a coil of effective area A, it carries a current I. The magnetic field B is radial. The torque acting on the coil is

$N A^{2} B^{2} I$

$N A B I^{2}$

$N^{2} A B I$

$N A B I$

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Explanation

(d) $τ = M B \sin θ \Rightarrow τ_{m a x} = N i A B, (θ = 90 °)$

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If a particle of charge $10^{- 12}$ coulomb moving along the $\hat{x} -$ direction with a velocity $10^{5}$ m/s experiences a force of $10_{}^{- 10}$ newton in $\hat{y}$ - direction due to magnetic field, then the minimum magnetic field is

$6.25 \times 10^{3}$ tesla in $\hat{z}$ - direction

$10^{- 15}$ tesla in $\hat{z}$ - direction

$6.25 \times 10^{- 3}$ tesla in $\hat{z}$ - direction

$10^{- 3}$ tesla in $\hat{z}$ - direction

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Explanation

(d) $F = qvB \sin θ \Rightarrow B = \frac{F}{qv \sin θ}$

$B_{\min} = \frac{F}{qv}$ (when $θ$ = $90 °$ )

$∴ B_{\min} = \frac{F}{qv} = \frac{10^{- 10}}{10^{- 12} \times 10^{5}} = 10^{- 3}$ Tesla in $\hat{z}$ - direction

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A small coil of N turns has an effective area A and carries a current I. It is suspended in a horizontal magnetic field $\dot{B}$ such that its plane is perpendicular to $\dot{B}$ . The work done in rotating it by $180 °$ about the vertical axis is

$N A I B$

$2 N A I B$

$2 πNAIB$

$4 π N A I B$

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Explanation

(b) $W = M B (\cos θ_{1} - \cos θ_{2})$

$= (N i A) B (\cos 0 ° - \cos 180 °) = 2 N A I B$

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An electron moves with a constant speed v along a circle of radius r. Its magnetic moment will be (e is the electron's charge)

evr

$\frac{1}{2}$ evr

${πr}^{2}$ ev

2 $π$ rev

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Explanation

(b) $M = i ({πr}^{2}) = \frac{e v}{2 πr} \times {πr}^{2} \Rightarrow M = \frac{1}{2} evr$

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A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{8} th$ to its value at the centre of the coil, is

$\frac{R}{\sqrt{3}}$

$R \sqrt{3}$

$2 \sqrt{3} R$

$\frac{2}{\sqrt{3}} R$

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Explanation

(b)

$\frac{B_{centre}}{B_{axis}} = {(1 + \frac{x^{2}}{R^{2}})}^{3 / 2}, also B_{axis} = \frac{1}{8} B_{centre} \Rightarrow \frac{8}{1} = {(1 + \frac{x^{2}}{R^{2}})}^{3 / 2} \Rightarrow 2 = {(1 + \frac{x^{2}}{R^{2}})}^{1 / 2} \Rightarrow 4 = 1 + \frac{x^{2}}{R^{2}} \Rightarrow 3 = \frac{x^{2}}{R^{2}} \Rightarrow x^{2} = 3 R^{2} \Rightarrow x = \sqrt{3} R$

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The sensitivity of a moving coil galvanometer can be increased by decreasing

The number of turns in the coil

The area of the coil

The magnetic field

The couple per unit twist of the suspension

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Explanation

(d) Sensitivity $S = \frac{θ}{i} = \frac{n B A}{C}$

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