# Class 12 Physics Chapter 4 Important Questions Moving Charges and Magnetism

Science is a complex and challenging subject, as it involves so many principles and concepts that are difficult to memorize. Those student who opt for science have to face many challenges and work hard to get good marks in the exam. In this lesson, students will learn about **Moving Charges and Magnetism**. The best solution of the problem is to practice as many Physics Class 12 Chapter 4 Important Questions as possible to clear the doubts.

**Q 1:-** Define one tesla (1 T) of magnetic field.
**Ans:-** Magnetic field is said to be one tesla if a unit positive charge moving in a direction perpendicular to field with a unit velocity of 1 m/s experiences a force of 1 N.
Alternately, magnetic field is said to be one tesla, if a straight conductor of unit length containing 1 A normal to the direction of field experiences a force of 1N.

Thus,

1 tesla `=\frac{1N}{1 A-m}`

`=1 N (A-m)^-1`

**Q 2:-** Name the physical quantity whose SI unit is Wb m^{-2}. Is it a scalar or a vector quantity?
**Ans:-** Magnetic field vector **B** has its SI unit Wb m^{-2}. (In fact, 1 tesla is also referred to as 1 Wb m^{-2}, i.e., IT = 1 Wb m^{-2}). Magnetic field is a vector.

**Q 3:-** Under what condition, dose an electron moving through a magnetic field experience maximum force?
**Ans:-** When an electron is moving in a direction perpendicular to that of magnetic field, it experiences a maximum force of magnitude, F= evB.

**Q 4:-** Under what condition is the force acting on a charge moving through a uniform magnetic field minimums?
**Ans:-** When a charge particle move parallel or antiparallel to the direction of a uniform magnetic field **B**, it does not experience any force (or experiences a minimum force having zero value).

**Q 5:-** An electron is moving along positive X- axis in the presence of uniform magnetic field along positive Y- axis. What is the direction of force acting on it?
**Ans:-** The force `F=-e(v\hat{i}\times\B\hat{j})=-evB(\hat{i}\times\hat{j})=-evB \hat{k}`
is along the negative Z- axis.

**Q 6:-** State Biot-Savart low in vector notation.
**Ans:-** `\vec{dB}=\frac{\mu_0}{4\pi}\frac{i\vec{dl}\times\vec{r}}{r^3}`

where i `\hat{dl}` is the current element and `\vec{r}` is the position vector of the point, where magnetic field `d\vec{B}` is being calculated, from the current element.

**Q 7:-** What is the CGS unit of magnetic field `\vec{B}`? How is it related to tesla?
**Ans:-** A gauss (1 G) is the CGS unit of magnetic field `\vec{B}`, where 1 G = 10^{-4} T.

**Read also:** Moving Charges and Magnetism Class 12 Physics Notes Chapter 4

**Q 8:-** State any rule which predicts the direction of magnetic field due to current flowing through a straight conductor.
**Ans:-** Right- hand thumb rule predicts the direction of magnetic field. As per this rule, hold the current- carrying straight conductor in your right- hand such that the thumb points along the direction of current, then the fingers will curl round the wire in the direction of magnetic field produced due to that current.

**Q 9:-** How does magnetic field at the centre of a circular coil carrying current depend on:

- (a) the radius of coil, and
- (b) the number of turns in the coil?

**Ans:-** Magnetic field developed at the centre of a circular coil carrying current:

- (a) is inversely proportional to its radius ( B ∝ 1/R), and
- (b) directly proportional to the number of turns in the coil (B ∝ N).

**Q 10:-** At a large distance x along the axis of a circular current- carrying coil ,how does field B depend on x?
**Ans:-** At a large distance x (x >> R) along the axis of a current- carrying circular coil, the magnetic field varies inversely as the cube of distance x, i.e., B ∝ 1/x^{3}.

**Q 11:-** Does the magnetic field produced at the centre of a solenoid coil, depend on

- (a) its diameter, and
- (b) its total length? Give reason.

**Ans:-** Magnetic field at the centre of a current-carrying solenoid is given by B = Î¼_{0}nI , where n is the turns per unit length, consequently, magnetic field B depends neither on the diameter nor or the total length of the solenoid.

**Q 12:-** What is the direction of magnetic field developed due to a circular current loop?
**Ans:-** Magnetic field due to a circular current loop is in a plane perpendicular to the current loop and the direction is given by right- hand thumb rule.

**Q 13:-** Obtain the dimensional formula for magnetic field vector_.
**Ans:-** Since tesla, the SI unit of magnetic field is given by

1 T = 1 N A^{-1}m^{-1}

Hence, dimensional formula for `\hat{B}`,

`[\vec{B}] =[MT^{-2}A^{-1}]`

**Q 14:-** What would be the path of a charged particle moving perpendicular to a uniform magnetic field?
**Ans:-** A circular path.

**Q 15:-** What is the force that a conductor Î”\vec{I}, carrying a current I, experiences when placed in a magnetic field \vec{B}? What is the direction of force?
**Ans:-** The force `Î”\vec{F}=I(Î”\vec{l}\times\vec{B})`. The force acts perpendicular to the plane containing `Î”\vec{l}` and `\vec{B}` in the direction of `(Î”\vec{l}\times\vec{B})` where direction of `Î”\vec{l}` is taken same as the direction of current I flowing through it.

**Read also:** Class 12 Physics Chapter 4 MCQs with Answer Moving Charges and Magnetism

**Q 16:-** What is the magnetic moment of a current loop? What is its direction?
**Ans:-** The magnetic moment of a current loop is given by `\vec{m}=i\vec{A}`. Thus, magnitude of magnetic moment is given by the product of current and the area of closed loop. The direction of magnetic moment `\vec{m}` is given by right-hand rule. As per this rule, hold your right hand so that the finger curl in the direction of electric current in the loop, then the thumb will give the direction of magnetic moment.

**Q 17:-** Does magnetic moment of a current loop depend on the shape of current loop?
**Ans:-** The magnetic moment of a current loop depends on the strength of current and the area of current loop. However, it does not depend on the shape of the current loop.

**Q 18:-** How does radius of the circular path traversed by a charged particle under the influence of a normal, uniform magnetic field depend on (a) its velocity and (b) its momentum, and (c) its kinetic energy?
**Ans:-** The radius r of the circular path of a charged particle of mass m and charge q under the influence of a normal magnetic field B depends on its speed v or momentum p or kinetic energy K.

As per the relation,

`r=\frac{mv}{qB}=\frac{p}{qB}=\frac{\sqrt{2mK}}{qB}`

**Q 19:-** What time does a charged particle take in completing a circular path in a uniform magnetic field? Does it depend on speed of the particle or the radius of the circular path?
**Ans:-** The time period (i.e., the revolution time) of a charged particle in a uniform magnetic field is given as:

`T =\frac{2\pi m}{qB}`

The time period is independent of the speed of the particle as well as the radius of the circular path.

**Q 20:-** A stream of electrons traveling with a speed v at right angles to uniform magnetic field B is deflected in a circular path of radius r.
Prove that e/m=v/rB.
**Ans:-** When a stream of electrons describe a uniform circular path under a normal magnetic field B, the magnetic force provides the requisite centripetal force.

Hence,

`\frac{mv^2}{r}=evB ⇒ \frac{e}{m}=\frac{v}{rB}`

**Q 21:-** State the condition for repeated acceleration of charged particle in a cyclotron.
**Ans:-** For continuous acceleration of a charged particle in a cyclotron, the oscillator frequency must coincide with frequency of natural revolution of charged particles.

**Q 22:-** What is the nature of force present between two parallel wire carring current same direction? What happens if one any one of the current is reversed?
**Ans:-** The force present between two parallel wires carrying current in the same direction is attractive in nature. If the direction of any one current is reversed, the force becomes repulsive in nature.

**Q 23:-** Estimate the torque on a closed current loop placed in a uniform magnetic field_?
**Ans:-** When a closed current loop of area A and carrying a current I is placed in a uniform magnetic field B, it experiences a torque, Ï„ = IAB sin Î¸, where Î¸ is the angle between the magnetic field and normal to the plane of current loop (i.e., Î¸ is the angle between **A** and **B**).

**Q 24:-** A coil of N turns, each turn of area A , carrying a current I is placed parallel to a uniform magnetic field B. What is the net force acting on the coil? What is the net torque acting on the coil?
**Ans:-** Net force acting on the coil is zero and the net torque is NAIB.

**Q 25:-** What is the principle of a moving coil galvanometer?
**Ans:-** When a current-carrying coil is placed in a uniform magnetic field, it experiences a torque. Under the influence of the influence of the torque, (i.e., deflection) is proportional to the current.

**Q 26:-** What is the nature of the magnetic field in a moving coil galvanometer coil.
**Ans:-** The magnetic field is a radial magnetic field whose direction is always perpendicular to the area vector of the galvanometer coil.

**Q 27:-** What is the function of radial magnetic field in a moving coil galvanometer?
**Ans:-** For a radial magnetic field **A** and **B** are normal to each other, hence torque acting on galvanometer coil, `\vec{Ï„}=\vec{m}\times\vec{B}=NI(\vec{A}\times\vec{B})` has a constant magnitude, Ï„ = NIAB for all orientations of the coil.

**Q 28:-** Why are pole pieces of horse shoe magnet used in to moving coil galvanometer?
**Ans:-** Pole pieces of magnet are made concave so as to make the magnetic field radial.

**Q 29:-** What is the main function of a soft iron core used in a moving coil galvanometer?
**Ans:-** The soft iron core used in a moving coil galvanometer:

- (a) helps in making the magnetic field radial, and
- (b) increases the strength of magnetic field.

**Q 30:-** Why should the spring in a moving coil galvanometer have low torsional constant?
**Ans:-** The spring used in a moving coil galvanometer has low torsional constant k so that galvanometer has higher value of current sensitivity and gives a large deflection even for an extremely weak current.

**Q 31:-** What is a shunt? What is its function?
**Ans:-** A shunt is a low resistance connected in parallel to an electrical circuit. Its function is to allow only a small fraction of the total current to pass through the given electrical circuit and allow the remaining current to pass through itself.

**Q 32:-** How can a galvanometer be converted into an ammeter?
**Ans:-** A galvanometer can be converted into an ammeter by connecting a shunt resistance of appropriate value in parallel to the galvanometer.

**Q 33:-** what is the resistance of an (a) ideal ammeter, and (b) ideal voltmeter?
**Ans:-** The resistance of an ideal ammeter is zero but that of an ideal voltmeter is infinite.

**Q 34:-** What is magnetic moment associated with the orbital motion of a charge q moving in a circle of radius r with a constant speed v?
**Ans:-** The magnetic moment,

`|\vec{m}| =\frac{qvr}{2}`

**Q 35:-** What is Bohr magneton? What is its value?
**Ans:-** Bohr magneton is the least possible value of magnetic moment associated with the orbital motion of an electron in an atom.
Value of Bohr magnetor Î¼B,

`\mu_B=\frac{eh}{4\pi m_e}=9.27\times10^{-24} Am^2`

**Q 36:-** Is a magnetic field set up around a stationary charge?
**Ans:-** No magnetic field is set up around a stationary charge. Magnetic field is associated with a moving charge only.

**Q 37:-** In a certain arrangement a proton or electron does not get deflected while lo passing through a magnetic field region. Under what condition is it possible?
**Ans:-** Force experienced by any charged particle q in a magnetic field region,

`\vec{F}=q(\vec{v}\times\vec{B})`

The charged particle goes undeflected i.e., does not experience any force only when `\vec{v}` and `\vec{B}` are either parallel or antiparallel.

**Q 38:-** A current is set up in a long copper pipe. Is there a magnetic field, and (a) inside (b) outside the pipe? Give reason.
**Ans:-**

- (a) At any point inside the current- carrying hollow copper pipe, magnetic field is zero Ampere's law because no current is being enclosed by the Amperion loop.
- (b) At any point outside the current pipe, there is a magnetic field whose strength falls inversely to the normal distance from the centre of pipe. Infact, field at a point situated at a distance r from the centre of current-carrying pipe is given by

`B=\frac{\mu_0I}{2\pi r}`

**Q 39:-** What will be the path of a charged particle moving along the direction of a uniform magnetic field?
**Ans:-** When a charged particle is moving along the direction of a uniform field, force acting on it `F=q(\vec{v}\times\vec{B})` has zero magnitude. In the absence of a magnetic force the charged particle continues to move along its original straight path without any change straight path without any change either in magnitude or in direction of its velocity.

**Q 40:-** An electron and a proton moving with the same speed enter the same magnetic field region at right angles to the direction of the field. For which of the two particles will the radius of circular path be smaller?
**Ans:-** From the relation, r = mv/qB, we know that all other factors remaining same r ∝ m. As out of electron and proton the mass of an electron is much less than of a proton, hence radius of circular path described by an electron is smaller.

**Q 41:-** What are the primary functions of electric field and the magnetic field in a cyclotron?
**Ans:-** In a cyclotron magnetic field makes the charged particle to cross the gap between the dees again and again by making it to move along a circular path.
The oscillating electric field applied across the dees accelerates the charged particle again by again in small steps so that final energy of charged particle is exceedingly high.

**Q 42:-** In what ways, electric and magnetic field differ?
**Ans:-** Electric field `\vec{E}` and magnetic field `\vec{B}` differ in following respects:

- (a) Electric field can be due to charge at rest or in motion but the magnetic field is basically due to moving charges (or currents) only.
- (b) Electric field lines in space do not form closed paths but magnetic field lines may have closed paths.
- (c) In the presence of a material medium, the magnitude of electric field decrease. However, magnetic field may increase in the presence of a ferromagnetic material medium.

**Q 43:-** What force does a charge placed near a current-carrying conductor experience, and why?
**Ans:-** A stationary charge placed near a current-carrying conductor does not experience any force. A current-carrying conductor produces a magnetic field around it. Charge experience a force in a magnetic field only when it is in motion.

**Q 44:-** When a charged particle moves in a magnetic field, its kinetic energy remains unchanged. Explain.
**Ans:-** The force on a moving charge,

`\vec{F}=q(\vec{v}\times\vec{B})`

is always perpendicular to `\hat{v}` and so the work done by the magnetic force is always zero. As a result, in accordance with work- kinetic energy principle, the kinetic energy of the charged particle remains unchanged.

**Q 45:-** Can a cyclotron accelerate neutrons? Give reason too.
**Ans:-** A cyclotron cannot be used to accelerate neutrons because neutrons are uncharged particles and do not electric or a magnetic field.

**Q 46:-** Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?
**Ans:-** Within the core of a current- carrying toroid coil the magnetic field lines make a closed path and are, thus, confined within the core itself. However, in a straight solenoid, magnetic field lines are along the axis of solenoid and escape from its two ends. Hence, field lines cannot be entirely confined within a straight solenoid.

**Q 47:-** Two identical charged particles moving with the same speed enter a region of uniform magnetic field B. If one of these enters normal to the field direction and the other enters along a direction of 30° with the field, what would be the ratio of their angular frequencies?
**Ans:-** As angular frequency Ï‰ = qB/m is independent of the angle of entrance of a charged particle in a magnetic field,

Hence, `Ï‰_1=Ï‰_2` or `\frac{Ï‰}{Ï‰}=1`

**Q 48:-** A current- carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis).
**Ans:-** No, it is not possible because for turning of loop about the vertical axis the torque Ï„ acting on current- carrying loop must be in the vertical direction. However, `\vec{Ï„}=I\vec{A}\times\vec{B}` and for a horizontal loop A must be in vertical direction. So, `(\vec{A}\times\vec{B})` would always be in the plane of the loop itself for any value of `\vec{B}`.

**Q 49:-** What is the effect in a vertical spring carrying a weight when a current is passed through the spring?
**Ans:-** As the current passes through the neighbouring loops of sping in the same direction, these will attract one another . As a result, the spring contracts and the weight moves upwards.

**Q 50:-** A charged particle is deflected either by an electric or a magnetic field. How can we ascertain the nature of the field?
**Ans:-** By observing the trajectory and measuring the kinetic energy of the charged particle we can ascertain the nature of field. In a magnetic field, the trajectory of charged particle is a circle in a plane perpendicular to the field and the speed and kinetic energy of charged particle remain constant. In an electric field, the trajectory of charged particle is a parabola in the plane of the field and speed and kinetic energy of particle change.

**Q 51:-** An HT electric power line is carrying current from west and east. Will the power line experience a force due to horizontal component B_{H} of the Earth's magnetic field? If yes, in what direction?
**Ans:-** The electric power line carries current from west and east along horizontal direction and horizontal component of earth's magnetic field is from south to north. As I`\vec{dl}` and `\vec{B_H}` are mutually perpendicular, the power line must experience a magnetic force due to `\vec{B_H}`. As per Fleming's left-hand thumb rule, the force is directed in the vertically upward direction.