CBSE 12th Standard Physics Subject Moving Charges And Magnetism Chapter Case Study Questions 2021
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CBSE 12th Standard Physics Subject Moving Charges And Magnetism Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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Various methods can be used to measure the mass of an atom. One possibility is through the use of a mass spectrometer. The basic feature of a Banbridge mass spectrometer is illustrated in figure. A particle carrying a charge +q is first sent through a velocity selector and comes out with velocity v = E/B.
The applied electric and magnetic fields satisfy the relation E = vB so that the trajectory of the particle is a straight line. Upon entering a region where a second magnetic field \(\vec{B}_{0}\) pointing into the page has been applied, the particle will move in a circular path with radius r and eventually strike the photographic plate.
(i) In mass spectrometer, the ions are sorted out in which of the following ways?(a) By accelerating them through electric field (b) By accelerating them through magnetic field (c) By accelerating them through electric and magnetic field (d) By applying a high voltage (ii) Radius of particle in second magnetic field Bo is
\(\text { (a) } \frac{2 m v}{q E_{0}}\) \(\text { (b) } \frac{m v}{q E_{0}}\) \(\text { (c) } \frac{m v}{q B_{0}}\) \(\text { (d) } \frac{2 m E_{0} v}{q B_{0}}\) (iii) Which of the following will trace a circular trajectory wit largest radius?
(a) Proton (b) -\(\alpha\)particle (c) Electron (d) A particle with charge twice and mass thrice that of electron (iv) Mass of the particle in terms q, Bo, B,r and E is
\(\text { (b) } \frac{q B_{0} B r}{E}\) \(\text { (c) } \frac{q B r}{E B_{0}}\) \(\text { (d) } \frac{q B r E}{B_{0}}\) (v) The particle comes out of velocity selector along a straight line, because
(a) electric force is less than magnetic force (b) electric force is greater than magnetic force (c) electric and magnetic force balance each other (d) can't say. -
An electron with speed Vo << c moves in a circle ofradius ro in a uniform magnetic field. This electron is able to traverse a circular path as magnetic field is perpendicular to the velocity of the electron. A force acts on the particle perpendicular to both \(\vec{v}_{0}\) and \(\vec{B}\). This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is To .
(i) If the speed of the electron is now doubled to 2vo.The radius of the circle will change to\(\text { (a) } 4 r_{0}\) \(\text { (b) } 2 r_{0}\) \(\text { (c) } r_{0}\) \(\text { (d) } r_{0} / 2\) (ii) If vo = 2vo then the time required for one revolution of the electron will change to
\(\text { (a) } 4 T_{0}\) \(\text { (b) } 2 T_{0}\) \(\text { (c) } T_{0}\) \(\text { (d) } T_{0} / 2\) (iii) A charged particles is projected in a magnetic field \(\vec{B}=(2 \hat{i}+4 \hat{j}) \times 10^{2} \mathrm{~T}\) The acceleration of the particle is found to be \(\vec{a}=(x \hat{i}+2 \hat{j}) \mathrm{m} \mathrm{s}^{-2}\). Find the value of x.
(a) 4 m S-2 (b) -4 m s-2 (c) -2 m s-2 (d) 2 m s-2 (iv) If the given electron has a velocity not perpendicular to B, then trajectory of the electron is
(a) straight line (b) circular (c) helical (d) zig-zag (v) If this electron of charge (e) is moving parallel to uniform magnetic field with constant velocity v, the force acting on the electron is
(a) Bev \(\text { (b) } \frac{B e}{v}\) \(\text { (c) } \frac{B}{e v}\) (d) zero -
Moving coil galvanometer operates on Permanent Magnet Moving Coil (PMMC) mechanism and was designed by the scientist D'arsonval.
Moving coil galvanometers are of two types
(i) Suspended coil
(ii) Pivoted coil type or tangent galvanometer.
Its working is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque. This torque tends to rotate the coil about its axis of suspension in such a way that the magnetic flux passing through the coil is maximum.
(i) A moving coil galvanometer is an instrument which(a) is used to measure emf (b) is used to measure potential difference (c) is used to measure resistance (d) is a deflection instrument which gives a deflection when a current flows through its coil (ii) To make the field radial in a moving coil galvanometer
(a) number of turns of coil is kept small (b) magnet is taken in the form of horse-shoe (c) poles are of very strong magnets (d) poles are cylindrically cut (iii) The deflection in a moving coil galvanometer is
(a) directly proportional to torsional constant of spring (b) directly proportional to the number of turns in the coil (c) inversely proportional to the area of the coil (d) inversely proportional to the current in the coil (iv) In a moving coil galvanometer, having a coil of N-turns of area A and carrying current I is placed in a radial field of strength B.
The torque acting on the coil is\(\text { (a) } N A^{2} B^{2} I\) \(\text { (b) } N A B I^{2}\) \(\text { (c) } N^{2} A B I\) (d) NABI (v) To increase the current sensitivity of a moving coil galvanometer, we should decrease
(a) strength of magnet (b) torsional constant of spring (c) number ofturns in coil (d) area of coil -
A galvanometer can be converted into voltmeter of given range by connecting a suitable resistance Rs in series with the galvanometer, whose value is given by
\(R_{s}=\frac{V}{I_{g}}-G\)
where V is the voltage to be measured, Ig is the current for full scale deflection of galvanometer and G is the resistance of galvanometer
Series resistort (Rs) increases range of voltmeter and the effective resistance of galvanometer. It also protects the galvanometer from damage due to large current.
Voltmeter is a high resistance instrument and it is always connected in parallel with the circuit element across which potential difference is to be measured. An ideal voltmeter has infinite resistance
In order to increase the range of voltmeter n times the value of resistance to be connected in series with galvanometer is Rs = (n - l)G.
(I) 10 mA current can pass through a galvanometer of resistance \(25 \Omega\) What resistance in series should be connected through it, so that it is converted into a voltmeter of 100 V?\(\text { (a) } 0.975 \Omega\) \(\text { (b) } 99.75 \Omega\) \(\text { (c) } 975 \Omega\) \(\text { (d) } 9975 \Omega \text { . }\) (ii) There are 3 voltmeter A, B, C having the same range but their resistance are \(15,000 \Omega, 10,000 \Omega\) and \(5,000 \Omega\) respectively. The best voltmeter amongst them is the one whose resistance is
\(\text { (a) } 5000 \Omega\) \(\text { (b) } 10,000 \Omega\) \(\text { (c) } 15,000 \Omega\) (d) all are equally good (iii) A milliammeter of range 0 to 25 mA and resistance of \(10 \Omega\) is to be converted into a voltmeter with a range of 0 to 25 V. The resistance that should be connected in series will be
(a) \(930 \Omega\) \(\text { (b) } 960 \Omega\) \(\text { (c) } 990 \Omega\) \(\text { (d) } 1010 \Omega\) (iv) To convert a moving coil galvanometer (MCG) into a voltmeter
(a) a high resistance R is connected in parallel with MCG (b) a low resistance R is connected in parallel with MCG (c) a low resistance R is connected in series with MCG (d) a high resistance R is connected in series with MCG (v) The resistance of an ideal voltmeter is
(a) zero (b) low (c) high (d) infinity -
A charged particle moving in a magnetic field experiences a force that is proportional to the strength of the magnetic field, the component of the velocity that is perpendicular to the magnetic field and the charge of the particle.
This force is given by \(\vec{F}=q(\vec{v} \times \vec{B})\) where q is the electric charge of the particle, v is the instantaneous velocity of the particle, and B is the magnetic field (in tesla).
The direction of force is determined by the rules of cross product of two vectors
Force is perpendicular to both velocity and magnetic field. Its direction is same as \(\vec{v} \times \vec{B}\) if q is positive and opposite of \(\vec{v} \times \vec{B}\) if q is negative
The force is always perpendicular to both the velocity of the particle and the magnetic field that created it. Because the magnetic force is always perpendicular to the motion, the magnetic field can do no work on an isolated charge. It can only do work indirectly, via the electric field generated by a changing magnetic field.
(I) When a magnetic field is applied on a stationary electron, it(a) remains stationary (b) spins about its own axis (c) moves in the direction of the field (d) moves perpendicular to the direction of the field. (ii) A proton is projected with a uniform velocity v along the axis of a current carrying solenoid, then
(a) the proton will be accelerated along the axis (b) the proton path will be circular about the axis (c) the proton moves along helical path (d) the proton will continue to move with velocity v along the axis. (iii) A charged particle experiences magnetic force in the presence of magnetic field. Which of the following statement is correct?
(a) The particle is stationary and magnetic field is perpendicular. (b) The particle is moving and magnetic field is perpendicular to the velocity (c) The particle is stationary and magnetic field is parallel (d) The particle is moving and magnetic field is parallel to velocity (iv) A charge q moves with a velocity 2 ms-1 along x-axis in a uniform magnetic field \(\vec{B}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{T}\) then charge will experience a force
(a) in z-y plane (b) along -yaxis (c) along +z axis (d) along -z axis (v) Moving charge will produce
(a) electric field only (b) magnetic field only (c) both electric and magnetic field (d) none ofthese.
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CBSE 12th Standard Physics Subject Moving Charges And Magnetism Case Study Questions 2021 Answer Keys
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(i) (c): In mass spectrometer, the ions are sorted out by accelerating them through electric and magnetic field.
(ii) (c): As \(\frac{m v^{2}}{r}=q v B_{0} \therefore r=\frac{m v}{q B_{0}}\)
(iii) (b): As radius \(r \propto \frac{m}{q}\)
\(\therefore\) r will be maximum for \(\alpha\) - particle.
(iv) (b) : Here, \(r=\frac{m v}{q B_{0}} \text { or } m=\frac{r q B_{0}}{v}\)
As \(v=\frac{E}{B}, \therefore m=\frac{q B_{0} B r}{E}\)
(v) (c): From the relation v = E/B, it is clear electric and magnetic force balance each other. -
(i) (b): As \(r_{0}=\frac{m v}{q B} \Rightarrow r^{\prime}=\frac{m\left(2 v_{0}\right)}{q B}=2 r_{0}\)
(ii) (c): As, \(T=\frac{2 \pi m}{q B}\)
Thus, it remains same as it is in dependent of velocity
(iii) (b): As \(F \perp B\)
Hence, \(a \perp B\)
\(\therefore \vec{a} \cdot \vec{B}=0\)
\(\Rightarrow \quad(x \hat{i}+2 \hat{j}) \cdot(2 \hat{i}+4 \hat{j})=0\)
\(2 x+8=0 \Rightarrow x=-4 \mathrm{~m} \mathrm{~s}^{-2}\)
(iv) (c): If the charged particle has a velocity not perpendicular to \(\vec{B},\) then component of velocity along \(\vec{B}\) remains unchanged as the motion along the \(\vec{B}\) will not be affected by \(\vec{B}\).
Then, the motion of the particle in a plane perpendicular to \(\vec{B}\) is as before circular one. Thereby, producing helical motion.
(v) (d): The force on electron \(F=q v B \sin \theta\)
As the electron is moving parallel to B
So,\(\theta=0^{\circ} \Rightarrow q v B \sin 0^{\circ}=0\) -
(I) (d): A moving coil galvanometer is a sensitive instrument which is used to measure a deflection when a current flows through its coil.
(ii) (d): Uniform field is made radial by cutting pole pieces cylindrically.
(iii) (b): The deflection in a moving coil galvanometer \(\phi=\frac{N A B}{k} \cdot I \text { or } \phi \propto N\) where Nis number of turns in a coil, B is magnetic field and A is area of cross-section.
(iv) (d): The deflecting torque acting on the coil
\(\tau_{\text {deflection }}=N I A B\)
(v) (b): Current sensitivity of galvanometer
\(\frac{\phi}{I}=S_{i}=\frac{N B A}{k}\)
Hence, to increase (current sensitivity) Si (torsional constant of spring) k must be decrease. -
(i) (d): A galvanometer can be converted into a voltmeter of given range by connecting a suitable high resistance R in series of galvanometer, which is given by
\(R=\frac{V}{I_{g}}-G=\frac{100}{10 \times 10^{-3}}-25=10000-25=9975 \Omega\)
(ii) (c): An ideal voltmeter should have a very high resistance.
(iii) (c): Resistance of voltmeter \(=\frac{25}{25 \times 10^{-3}}=1000 \Omega\)
\(\therefore \quad X=1000-10=990 \Omega\)
(iv) (d): To convert a moving coil galvanometer into a voltmeter, it is connected with a high resistance in series. The voltmeter is connected in parallel to measure the potential difference. As the resistance is high, the voltmeter itself does not consume current.
(v) (d): The resistance of an ideal voltmeter is infinity. -
(i) (a): For stationary electron, \(\vec{v}=0\)
\(\therefore\) Force on the electron is \(\vec{F}_{m}=-e(\vec{v} \times \vec{B})=0\)
(ii) (d): Force on the proton \(\vec{F}_{B}=e(\vec{v} \times \vec{B})\)
Since, \(\vec{v}\) is parallel to \(\vec{B}\)
\(\therefore \quad \vec{F}_{B} \doteq 0\)
Hence proton will continue to move with velocity v along the axis of solenoid.
(iii) (b): Magnetic force on the charged particle q is
\(\vec{F}_{m}=q(\vec{v} \times \vec{B}) \text { or } F_{m}=q v B \sin \theta\)
where \(\theta\) is the angle between \(\vec{v} \text { and } \vec{B}\)
Out of the given cases, only in case (b) it will experience the force while in other cases it will experience no force
(iv) (a) : \(\vec{F}=q(\vec{v} \times \vec{B})\)
\(=q[(2 \hat{i} \times(\hat{i}+2 \hat{j}+3 \hat{k})]=(4 q) \hat{k}-(6 q) \hat{j}\)
(v) (c): When an electric charge is moving both electric and magnetic fields are produced, whereas a static charge produces only electric field.
