Moving Charges and Magnetism Test - 19

As we know that the uniqueness of helical path is determined by its pitch

P(Pitch) = \(2\pi\, mv\,cos\theta\over Bq\)

Where \(\theta\) is angle of velocity of charge particle with x-axis

For the given pitch P corresponding to charge particle, we have

\({q\over m}={2\pi v\,cos\theta\over BP}\) = constant

If motion is not helical, (\(\theta\) = 0)

As charged particles traverse identical helical paths in a completely opposite direction in a same magnetic field B, LHS for two particles should be same and of opposite sign.

\(\therefore\) \(({e\over m})_1+({e\over m})_2=0\)

By Biot-Savart law

dB = \({IdlSin\theta\over r^2}\) = \(\Big({I\times dl\over r}\Big)\)

In Biot-Savat’s law, magnetic field B || idl x r and idl due to flow of electron is in opposite direction of v and by direction of cross product of two vectors \(B\perp V\) So, the magnetic field is \(\perp\) to the direction of flow of charge.

As the direction of magnetic moment of circular loop of radius R placed in the x-y plane is along z-direction and given by M = I\((\pi r^2)\), when half of the loop with x > 0 is now bent so that it now lies in the y-z plane, the magnitudes of magnetic field moment of each semicircular loop of radius R lie in the x-y plane and the y-z plane is M' = I\((\pi r^2)\)/4 and the direction of magnetic field moments are along z-directlon and x-direction respectively.

Then resultant is:

\(M_{net}=\sqrt{M'^2+M'^2}\) = \(\sqrt2 M'\) = \(\sqrt2I(\pi r^2)/4\)

So, \(M_{net}\) < M or M diminishes.

Hence, the magnitude of magnetic moment now diminishes.

Magnetic Lorentz force:

F = qVB sin\(\theta\)

Magnetic Lorentz force electron is projected with uniform velocity along the axis of a current carrying long solenoid F = –qvB sin 0° = 0 and electric field is zero (E = 0). Due to this, magnetic field (B) is perpendicular to the direction of motion (v). So it will not affect the velocity of moving charge particle. So the electron will continue to move with uniform velocity along the axis of the solenoid

There is crossed electric and magnetic field between dees so the charged particle accelerates by electric field between dees towards other dees. So, the charged particle undergoes acceleration as

(i) speeds up between the dees because of the oscillating electric field.

(ii) speed remain the same inside the dees because of the magnetic field but direction undergoes change continuously.

Hence, the charge particle accelerates inside and between Dees always

When the axis of rotation of the loop is along B and rotating by 30° about an axis perpendicular to its plane make no change in the angle made by axis of the loop with the direction of magnetic field. So, the work done to rotate the loop is zero.

W = MB cos 90° = 0

When rotation not along B.

The work done to rotate the loop in uniform magnetic field W = MB\((cos\theta_1-cos\theta_2)\), where signs are as usual. So the workdone is

W = MB cos \(\theta\) (\(\theta\) = 30° given)

= MB cos 30° = \(MB\sqrt3\over2\).

The gyro-magnetic ratio of an electron in an H-atom is equal to the ratio of the magnetic moment and the angular momentum of the electron

\(\mu_e={magnetic\,moment\, of \,e(M_e)\over angular\,momentum\, of \,l(L_e)}\)

If I is the magnitude of the angular momentum of the electron about the central nucleus (orbital angular momentum). Vectorially,

\(\mu_e={-evr\over 2m_evr}\)  \((\because\, M_e={-evr\over2},L_e=m_evr)\)

\(\mu_e={-e\over 2m_e}\)

So it is independent of velocity or orbit of e depends only on charge.

The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.

Magnetic forces on a current carrying conductor, consider wire carrying a steady current I, placed in a uniform magnetic field B, perpendicular to its length is

\( F = IlBsin90° = IlB\)

The direction of force is get by Fleming’s left hand rule and F is perpendicular to the direction of magnetic field B. Hence, work done by the magnetic force on the ions is 0.

Loops are identical placed coaxially and carrying same current in opposite sense.

By Ampere’s circuital law, we get

\(\oint B.dl=\mu_0(I-I)=0\)

As the magnetic field inside the loop, there may be a point on C where B and dl are perpendicular to the direction of plane of loop.

So, \(\oint B.dl=0\)

or, \(|B.d\vec l|\) cos90° = 0.

The magnetic forces on a charge particle, F = q(v x B), is either zero or F is perpendicular to V velocity and magnetic field, which in turn revolves particles on circular path perpendicular to both \((\vec B,\vec V)\) with uniform speed. In both the cases magnetic force on charge particles have equal accelerations.

Both the particles gain or loss energy at the same rate as both are subjected to the same electric force (F = qE) in opposite direction, because magnitudes of charges are constant.

As, there is no change of the centre of Mass of the charged particles. Hence, the motion of the Centre of Mass is determined by B alone.

The Lorentz force is experienced by the single moving charge in space is filled with some uniform electric and magnetic fields is

\(F_L=(F_E+F_m)\)

\(F_L=qE+q(v\times B)\)

The, force on charged particle due to electric field \(F_E\) = qE

Force on charged particle due to magnetic field \(F_m\) = q(v x B)

The velocity V of charge particle q in magnetic field (B) and electric field (E) will be constant. If Lorentz force \((F_L)\) = qE + q(v x B) = \( ((F_E) + (F_m))\) on charge q.

Now, \(F_E\) = 0 if E = 0 and \(F_m\) = 0 if sin\(\theta\) = 0 or \(\theta\)° = 0° or 180°. Hence, B \(\neq\) 0

Also, E = 0 and B = 0 and the resultant force qE + q (v x B) = 0. In this case, it may also happen that E \(\neq\) 0, B \(\neq\) 0.