Solution:

Figure of merit is defined as the amount of current which produces unit deflection in the galvanometer.

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Solution:

Current Sensitivity: It is defined as the deflection produced in a coil per unit current passed through it.

`I_s = alpha/I = (NBA)/k`

`I = (BAN)/K ;` if `N -> 2N `, the `l -> 2 l` and `R -> 2 R`

` :. V_s = (BAN)/(K R) = (BA (2N))/(K (2R)) = (BAN)/(K R)`

i.e. if the current sensitivity is doubled, say by doubling the number of turns, then the voltage sensitivity may not be increased because it will increase the resistance of the galvanometer and the voltage sensitivity may remain the same.

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Solution:

As `F_text(max) = qvB => B = F_text(max) /(qv)`


If `F_text( max) = 1 N, q = + 1 C` and `v = 1 m//s` and `theta = 90°`, then `B = 1 T`

Hence, one tesla is the magnetic field in which a normally entering `+ 1 C`

charge, moving at `1 ms^(-1)` experiences a maximum force of `1 N`.

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Solution:

It is the ratio of the magnetic dipole moment to the angular momentum of the electron revolving round the nucleus.

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Solution:

Current Sensitivity: It is defined as the deflection produced in a coil per unit current passed through it.

`I_s = alpha/I = (NBA)/k`

`I = (BAN)/K ;` if `N -> 2N `, the `l -> 2 l` and `R -> 2 R`

` :. V_s = (BAN)/(K R) = (BA (2N))/(K (2R)) = (BAN)/(K R)`

i.e. if the current sensitivity is doubled, say by doubling the number of turns, then the voltage sensitivity may not be increased because it will increase the resistance of the galvanometer and the voltage sensitivity may remain the same.

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Solution:

Consider an electron revolving around the nucleus of an atom. The electron is in a uniform circular motion around the nucleus of charge `+Ze`. This constitutes a current.

`:. I = e/T ` ............(i)

If r is the orbital radius of the electron and 'v' is the orbital speed, then the time period is given as

` T = (2 pi r)/v` .............(ii)

From equations (i) and (ii), we get

` I = (ev)/(2 pi r)` .............(iii)

As the magnetic moment is given by
`mu_l = I pi r^2`

We have ` mu_l = ( (ev)/(2 pi r)) pi r^2 = (evr)/2`

Multiplying and dividing by me in above equation, we get

`mu_l = (evm_e r)/( 2 m_e) = e/(2 m_e) l`

where `l` is the angular momentum of the electron. According to the Bohr hypothesis, the angular momentum can have discrete values only.

i.e ` l = (nh)/(2 pi)`

So, ` mu_l = (enh)/(2 pi ( 2 m)) = ( n e h)/(4 pi m)`

In this case it is directed into the plane of the paper.

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Solution:

Lorentz magnetic force, `vec F = q ( vec v xx vec B)`

` ∵` Work done `= vec F . vec S. (∵ vec S || vec v)`

As the force is acting in a direction normal to the velocity of a charged particle (i.e. displacement), no work is done by the force.

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Solution:

Lorentz magnetic force `( vec F_m ) = q ( vec v xx vec B)`. The direction of magnetic force is perpendicular to the plane containing velocity and magnetic field vectors.

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Solution:

Figure of merit is defined as the amount of current which produces unit deflection in the galvanometer.

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Solution:

Velocity selector is a device consisting of perpendicular electric and magnetic fields that can be used as a velocity filter for charged particles. It is used to measure charge to mass ratio and is also used in Mass Spectrometer.

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Solution:

`vec M = I vec A`, as vector `vec A` is perpendicular to the surface, the magnetic moment `vec M` will also be perpendicular to the plane of circular coil.

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