(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, \[{{B}_{E}}=\frac{{{H}_{E}}}{\cos I}\] \[=\frac{0.4\times {{10}^{-4}}}{\cos 60{}^\circ }=0.8\times {{10}^{-4}}T\]
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