question_answer

(i) The coil area of a galvanometer is \[25\times {{10}^{-4}}{{m}^{2}}.\] It consists of 150 turns of a wire and is in a magnetic field of 0.15 T. The restoring torque constant of the suspension fibre is \[{{10}^{-6}}N-m\] per degree.

Assuming the magnetic field to be radial, calculate the maximum current that can be measured by the galvanometer, if the scale can accommodate. \[30{}^\circ \] deflection.

(ii) Show that the electron revolving around the nucleus in a radius r with orbital speed v has magnetic moment evr/2. Write an expression of magnetic moment of the electron in terms of its angular momentum.

Or

(i) A bar magnet of magnetic moment \[1.5\text{ }J{{T}^{-1}}\] is aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work required to turn the magnet, so as to align its magnetic moment

(A) normal to the field direction?

(B) opposite to the field direction?

(b) What is the torque on the magnet in cases (A) and (B).

(ii) How does the angle of dip vary as one moves from the equator towards the North pole? If the horizontal component of Earth's magnetic field at a place when the angle of dip is \[60{}^\circ \] is \[0.4\times {{10}^{-\,4}}T.\] Calculate the vertical component and the resultant magnetic field of the Earth at that point.

Answer:

            (i) Given, coil area,         \[A=25\times {{10}^{-4}}\text{ }{{m}^{2}}\] Number of turns, N = 150, B = 0.15 T Restoring torque constant, \[k={{10}^{-6}}\text{ }N-m/degree\] Twist produced,  \[{{\phi }_{\max }}=30{}^\circ \] As total restoring torque produced \[=k\theta \] In equilibrium position of the coil, Deflecting torque = Restoring torque \[k{{\phi }_{\max }}=N{{I}_{\max }}AB\] \[\Rightarrow \]   \[{{I}_{\max }}=\frac{K{{\phi }_{\max }}}{NAB}\]             \[=\frac{{{10}^{-6}}\times 30}{150\times 25\times {{10}^{-4}}\times 0.15}\]             \[=0.053\times {{10}^{-2}}\]             \[5.3\times {{10}^{-4}}A\] (ii) Magnetic Dipole Moment of a Revolving Electron An electron being a charged particle, constitutes a current while moving in its circular orbit around the nucleus (\[\because \]moving charge constitutes a current as well as magnetic field). If T is the time period of revolution, then current constituted by electron is                         \[I=e/T\]                        ?(i) where, e = charge of electron. If r is the orbital radius of electron and its orbital speed is v, then \[T=\frac{2\pi r}{v}\] \[\Rightarrow \]   \[I=\frac{e}{2\pi /v}=\frac{ev}{2\pi r}\]  [From Eq. (i)] Magnetic moment of revolving electron, M = IA             \[M=\frac{ev}{2\pi r}\pi {{r}^{2}}\]             \[M=\frac{evr}{2}\] This can be written as, \[M=\frac{e}{2{{m}_{e}}}({{m}_{e}}vr)\] [where, me is the mass of electrons]                         \[M=\frac{e}{2{{m}_{e}}}l\] where, l is the angular momentum of the electron. Or (i) \[W=mB(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})\] \[\therefore \tau =m\times B=mB\sin \theta \] (a) (A) Given, m =1.5 J/T, B = 0.22 T, \[{{\theta }_{1}}=0{}^\circ ,\]             \[{{\theta }_{2}}=90{}^\circ \] \[\therefore \]      \[W=1.5\times 0.22(cos0{}^\circ -cos90{}^\circ )\]             = 0.33 J (B) \[\because \]  \[{{\theta }_{2}}=180{}^\circ \] \[\therefore \]      \[W=1.5\times 0.22\times (\cos 0{}^\circ -\cos 180{}^\circ )\]             = 0.66 J (b) \[\tau =mB\sin \theta \] (A) \[\theta =90{}^\circ \] \[\tau =1.5\times 0.22\times \sin 90{}^\circ \]             \[=1.5\times 0.22\times 1\]             \[=0.33N-m\] (B) \[\because \theta =180{}^\circ \] \[\tau =1.5\times 0.22\times \sin 180{}^\circ \] \[=1.5\times 0.22\times 0=0\,N-m\] (ii) Angle of dip,  \[I=60{}^\circ \]             \[{{H}_{k}}=0.4\times {{10}^{-4}}T\]             \[{{H}_{E}}={{B}_{E}}\cos I\]             \[{{Z}_{E}}={{B}_{E}}\sin I\] \[\therefore \]      \[\operatorname{tanI}=\frac{{{Z}_{E}}}{{{H}_{E}}}\] Vertical component, \[{{Z}_{E}}={{H}_{E}}\tan I\]                         \[=0.4\times {{10}^{-4}}\times \tan 60{}^\circ \]                         \[=6.9\times {{10}^{-5}}T\] \[\therefore \]      Resultant, \[{{B}_{E}}=\frac{{{H}_{E}}}{\cos I}\]                         \[=\frac{0.4\times {{10}^{-4}}}{\cos 60{}^\circ }=0.8\times {{10}^{-4}}T\]