624 + Interview Questions in MAGNETIC-EFFECT-OF-CURRENT IN GENERAL AWARENESS Page 2 InterviewSolution

51.	Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons `(`1 Bohr magneton equals `9.27 xx 10^(-24) A m^2``)`. The density of atoms in iron is `8.52 xx 10^28` atoms `m^(-3)`. (a) Find the maximum magnetization `I` in a long cylinder of iron. (b) Find the maximum magnetic field `B` on the axis inside the cylinder.
Answer» Correct Answer - `158xx10^6Am^-1`; `1985T` Here, magnetic moment of each atom `=185xx10^(-23)Am^2` Number of atoms/volume `=852xx10^(28)m^-3` `:.` Max. magnetisation, `I=M/V=185xx10^(-23)xx852xx10^(28)` `=158xx10^6Am^-1` Max. mag. Induction `B=mu_0(H+I)` `=4pixx10^-7(0+158xx10^6)=1*985T`

Discussion

52.	A `30cm` wire carrying a current of `10A` is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be `027T`. What is the magnetic force on the wire?
Answer» Here, `l=30cm=3xx10^-2m`, `I=10A`, `B=027T`, `theta=90^@`, `F=?` `F=Bil sin theta=027xx10xx(3xx10^-2)xxsin90^@=027xx10xx3xx10^-2xx1=8*1xx10^-2N` The direction of force is perpendicular to the direction of current as well as of magnetic field.

Discussion

53.	A circular segment of radius `20cm` subtends an angle of `60^@` at its centre. Figure A current of `10A` is flowing through it. Find the magnitude and direction of the magnetic field produced at the centre.
Answer» Correct Answer - `52xx10^-6T` Let `hatn` be a unit vector perpendicular to the plane of paper upwards. Then total magnetic field at O due to current through the given structur is `vecB=vecB_(ab)+vecB_(bc)+vecB_(cd)=0+(mu_0)/(4pi)I/rtheta(-hatn)+0` [As point O lies on the wire ab and cd, so `B_(ab)=0=B_(cd)`] `:. B=(mu_0)/(4pi)I/rtheta=10^-7xx(10)/((20xx10^-2))xxpi/3` `=52xx10^-6T`

Discussion

54.	A wire loop consists of a stright segment `AB` and a circular are `ACB` of radius r The segment `AB` subtends an angle of `60^(@)` at the centre `O` of the circular are The wire loop carries a current I in the clockwise direction The direction of the magnetic field `B` is .A. parallel to the plane of the coilB. perpendicular to the plane of the coil and directed out of the page .C. perpendicular to the plane of the coil and directed into the pageD. inclined at an angle of `60^(@)` with the plane of the coil .
Answer» Correct Answer - `C`

Discussion

55.	A wire loop consists of a stright segment `AB` and a circular are `ACB` of radius r The segment `AB` subtends an angle of `60^(@)` at the centre `O` of the circular are The wire loop carries a current I in the clockwise direction The net magnetic filed `B` at `O` due to the whole wire loop is .A. `B=B_(1)+B_(2)`B. `B =B_(2)-B_(1)`C. `B =sqrt(B_(1)^(2)+B_(2)^(2))`D. `B =sqrt(B_(2)^(2)+B_(1)^(2))`
Answer» Correct Answer - `A`

Discussion

56.	Write two factors by which the current sensitivity of a moving coil galvanometer can be increased?
Answer» A current sensitivity of moving coil galvanometer is `I_s=theta/I=(nBA)/(k)`. Thus, current sensitivity can be increased (i) by increasing the strength of magnetic field B or increasing the number of turns in the coil. (ii) by decreasing the torsional constant k of its suspension fibre.

Discussion

57.	Define the current sensitivity of a moving coil galvanometer and state its SI unit?
Answer» The current sensitivity of a moving coil galvanometer is defined as the deflection produced in the galvanometer when unit current flows through it. Current sensitivity, `I_s=theta/I=(nBA)/(k)` The SI unit of current sensitivity is `rad A^-1`.

Discussion

58.	In order to increase the sensitivity of a moving coil galvanometer, one should decreaseA. The strength of its magnB. The torsional constant of its suspensionC. The number of turns in its coilD. The area of its coil
Answer» Correct Answer - B

Discussion

59.	Four wires each of length `2.0` meters area bent into four loops `P,Q,R` and `S` and then suspended into uniform magnetic field. Same current is passed in each loop. Which statement is correct? A. Couple on loop `P` will be the highestB. Couple on loop `Q` will be the highestC. Couple on loop `R` will be the highestD. Couple on loop `S` will be the highest
Answer» Correct Answer - D

Discussion

60.	A proton moving with a constant velocity passes through a region of space without any changing its velocity. If `E` and `B` represent the electric and magnetic fields, respectively. Then, this region of space may haveA. `E=0, B=0`B. `E=0,B!=0`C. `E!=0,B=0`D. `E!=0,B!=0`
Answer» Correct Answer - A::B::D

Discussion

61.	Magnetic field at the centre of coil of `n` turns, bent in the form of a square of side `2l`, carrying current `i`, isA. `(sqrt(2)mu_(0)ni)/(pil)`B. `(sqrt(2)mu_(0)ni)/(2pil)`C. `(sqrt(2)mu_(0)ni)/(4pil)`D. `(2mu_(0)ni)/(pil)`
Answer» Correct Answer - A

Discussion

62.	Find the experssion for the magnetic field at the centre O of a coil bent in the form of a square of side `2a`, carrying current I, figure.
Answer» Here, `OE=AE=a`, `/_AOE=/_BOE=45^@` Magnetic field at O due to current in arm AB of wire `B_1=(mu_0)/(4pi)I/a[sin 45^@+sin45^@]=(sqrt2mu_0I)/(4pia)` It is acting vertically downwards. The magnetic field at O due to currents in arms BC, CD and DA will also be of the same magnitude and direction as `vecB_1`. Therefore, resultant magnetic field at O is `B=4B_1=(4xxsqrt2mu_0I)/(4pia)=(sqrtmu_0I)/(pia)` It is acting vertically downwards.

Discussion

63.	A thin wire of length `l` is carrying a constant current. The wire is bent to form a circular coil. If radius of the coil, thus formed, is equal to R and number of turns in it is equal to `n`, then which of the following graphs represent (s) variation of magnetic field induction (B) at centre of the coilA. B. C. D.
Answer» Correct Answer - B::C

Discussion

64.	A part of a long wire carrying a current `i` is bent into a circle of radius `r` as shown in figure. The net magnetic field at the centre `O` of the circular loop is A. `(mu_(0)i)/(4r)`B. `(mu_(0)i)/(2r)`C. `(mu_(0)i)/(2pir) (pi+1)`D. `(mu_(0)i)/(2pir)(pi-1)`
Answer» Correct Answer - C

Discussion

65.	Statement 1: A magnetic field independent of time can change the velocity of a charged particle. Statement 2: It is not possible to change the velocity of a particle in a magnetic field as magnetic field does no work on the charged particle.
Answer» Correct Answer - `C` Velocity is a vector quantity even if direction changes, velocity is said to be changing no matter speed remains same or different .

Discussion

66.	Two identical charged particles moving with same speed enter a region of uniform magnetic field. If one of these enters normal to the field direction and the other enters along a direction at `30^@` with the field, what would be the ratio of their angular frequencies?
Answer» When a charged particle of charge q, mass m, moving with velocity v enters a uniform magnetic field B, acting perpendicular to the direction of motion of charged particle, then the magnetic force on the particle provides centripetal force. So `qvB=mv^2//r` or `v//r=qB//m` The particle will describe a circular path of radius r. Angular frequency of particle, `omega=v/r=(qB)/(m)` When the charged particle is moving, making an angle `30^@` with the direction of magnetic field, then due to component velocity `v cos 30^@` (along the direction of field) no force acts on the particle. Due to component velocity ` v sin 30^@`, (acting normally to the direction of field ) of force is acting on the particle which provides the required centripetal force for circular motion of particle. Thus, `q(v sin 30^@)B=(m(v sin 30^@)^2)/(r)` or `(vsin 30^@)/(r)=(qB)/(r)` Angular frequency `omega=(v sin 30^@)/(r)=(qB)/(r)` Thus, the angular frequency of both the particles is same. So ratio of their angular frequency `=1:1`

Discussion

67.	What is sure test of magnetism?
Answer» Repulsion is sure test of magnetism.

Discussion

68.	A charged particle enters a (non-uniform) magnetic field varrying from point to point, both in magnitude and direction, with a certain initial velocity. What do you say about the final velocity of the particle when it leaves the field?
Answer» The force on the moving charged particle in magnetic field is `vecF=q(vecvxxvecB)`. So `vecF` is always perpendicular to the particle velocity `vecv`. Hence the magnitude of velocity remains unchanged, only the direction changes. Thus the particle will leave the varrying magnetic field with its initial velocity, but in a different direction.

Discussion

69.	`ABCD` is a square loop made of a uniform conducting wire. A current enters the loop at `A` and leaves at `D`. The magnetic field is A. Maximum at the centre of the loopB. Zero at the centre of loopC. Zero at all points inside the loopD. Zero at all points outside of the loop
Answer» Correct Answer - B

Discussion

70.	A magnetic fieldA. Always exerts a force on a charged particleB. Never exerts a force on a charged particleC. Exerts a force, if the charged particle is moving across the magnetic field linesD. Exerts a force, if the charged particle is moving along the magnetic field lines
Answer» Correct Answer - C

Discussion

71.	Write the relation for the force `vecF` acting on a charge carrier q moving with a velocity `vec` through a magnetic field `vecB` in vector notation. Using the reltation, deduce the conditions under which this force will be (i) maximum (ii) minimum.
Answer» `vecF=q(vecvxxvecB)` or `\|vecF\|=q\|vecvxxvecB\|=qvB sin theta` (i) F will be maximum, when `sin theta=1` or `theta=90^@`, i.e., the charged particle is moving perpendicular to the direction of magnetic field. (ii) F will be minimum, When `sin theta=0` or `theta=0^@` or `180^@` i.e., the charged particle is moving parallel to the direction of magnetic field.

Discussion

72.	The magnetic force acting on a charged particle of charge `-2 muC` in a magnetic field of `2 T` acting `y` direction, when the particle velocity is `(2i + 3 hat(j)) xx 10^(6) ms^(-1)`, isA. 8 N in z-directionB. 8 N in -z-directionC. 4N in z-directionD. 8N in y-direction
Answer» Here `q=-2muC=-2xx10^-6C`, `vecB=2hatjT`, `vecv=(2hati+3hatj)xx10^6ms^-1` Now force on charged particle, `vecF=q(vecvxxvecB)` `=-2xx10^-6[(2hati+3hatj)xx10^6xx2hatj]` `=-2xx10^-6xx10^6[(2hati+3hatj)xx(2hatj)]` `=-2[4hatk][ As hatixxhatj=hatk, hatjxxhatj=0]` `=-8hatk` [ 8N along -Z direction]

Discussion

73.	The magnetic force acting on a charged particle of charge `-2 muC` in a magnetic field of `2 T` acting `y` direction, when the particle velocity is `(2i + 3 hat(j)) xx 10^(6) ms^(-1)`, isA. 8 N in z-directionB. 4 N in z-directionC. 8 N in y-directionD. 8 N in z-direction
Answer» Correct Answer - a

Discussion

74.	The magnetic force acting on a charged particle of charge `-2 muC` in a magnetic field of `2 T` acting `y` direction, when the particle velocity is `(2i + 3 hat(j)) xx 10^(6) ms^(-1)`, isA. 8 N in z-directionB. 4 N in z-directionC. 8 N in y-directionD. 8 N in z-direction
Answer» Correct Answer - a

Discussion

75.	A square current carrying loop is suspended in a unifrom magnetic field acting in the palne of the loop. If the force on one arm of the loop is `vec(F)`, the net force on the remaining three arms of the loop isA. 3FB. `-F`C. `-3F`D. F
Answer» Correct Answer - b

Discussion

76.	A dip circle shows an apparent dip of `60^@` at a place where true dip is `45^@`. If dip circle is rotated through `90^@`, what apparent value of dip will it show?
Answer» Correct Answer - `51^@` Here, `delta_1=60^@`, `delta=45^@`, `delta_2=?` As `cot^2delta_1=cot^2delta_2=cot^2delta` `:. cot^2delta_2=cot^2delta-cot^2delta_1` `=(cot45^@)^2-(cot60^@)^2` `=1-(1/sqrt3)^2=2/3=067` `cot delta_2=sqrt(067)=0816` `:. delta_2=cot^-1(0816)=51^@`

Discussion

77.	If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity `v` and and electron is projected along the line of force, what will happen to proton and electron?A. The electron will travel along a circle with constant speed andthe proton will move along a straight lineB. Proton will move in a circle with constant speed and there willbe no effect on the motion of electronC. There will not be any effect on the motion of electron andprotonD. The electron and proton both will follow the path of a parabola
Answer» Correct Answer - B

Discussion

78.	A proton of energy `8 eV` is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will beA. `4eV`B. `2eV`C. `8eV`D. `6eV`
Answer» Correct Answer - C

Discussion

79.	A proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to direction of magnetic field. If both the particles move around magnetic field in circles of equal radii, the ratio of momentum of proton to alpha particle `(p_(p)/(p_(alpha)))` isA. 1B. `1/2`C. 2D. `1/4`
Answer» Correct Answer - B

Discussion

80.	An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively `R_(e), R_(p), R_(d)` and `R_(alpha)` It follows thatA. `R_(e)=R_(p)`B. `R_(p)=R_(d)`C. `R_(d)=R_(alpha)`D. `R_(p)=R_(alpha)`
Answer» Correct Answer - C

Discussion

81.	A proton, a deuteron and an `alpha` particle are accelerated through same potential difference and then they enter a normal uniform magnetic field, the ratio of their kinetic energies will beA. `1:2:2`B. `2:2:1`C. `1:2:1`D. `1:1:2`
Answer» Correct Answer - D

Discussion

82.	Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii `R_1 and R_2,` respectively. The ratio of masses of X and Y isA. `((R_(1))/(R_(2)))^(1//2)`B. `(R_(2))/(R_(1))`C. `((R_(1))/(R_(2)))^(2)`D. `(R_(1))/(R_(2))`
Answer» Correct Answer - C

Discussion

83.	A long thin wire is placed along the y-axis (just outside) of a frame of refernce. There exists a uniform magnetic field of `10^(-6)` T along the x-axis. Calculate the magnetic field at the points (0,0,2m),(0,2m,0) and (2m,0,0) when the wire carries 10A current. Revise the calculation for a thick wire.
Answer» Correct Answer - `2 xx 10^(-6)`T, `infity`, `sqrt(2) xx 10^(-6)`T, `2 xx 10^(-6)`T, `10^(-6)`T, `sqrt(2)xx 10^(-6)`T

Discussion

84.	A proton and ` alpha` - particle are accelerated with same potential difference and they enter in the region of constant magnetic field `B` perpendicular to the velocity of particles. Find the ratio of radius of curvature of ` alpha` - particle`.
Answer» Correct Answer - `(r_(p))/(r_(alpha)) = sqrt((m_(p)/(m_(alpha)).(q_(alpha))/(q_(p))) = (1)/(sqrt2)` `r = (mv)/(qB) = (p)/(qB) = sqrt(2mE)/(qB) = (sqrt2mqV)/(qB)` `(r_(p))/(r_(a)) = sqrt((m_(p))/(m_alpha) xx (q_alpha)/(q_p)) = sqrt((1)/(4) xx (2)/(1)) = (1)/(sqrt2)` .

Discussion

85.	Two long, straight parallel conductors each carrying 2A current in the same direction are 10 cm apart. Calculate the magnetic field at a distance a) at a point which is midway between the wires b) at a point which is 15 cm from one wire and 5cm from the other.
Answer» Correct Answer - a) Zero, b) `1.07 xx 10^(-5)`T

Discussion

86.	Calculate the magnetic field at the corner of a right angled triangle ABC, A being `90^(@)`, when two long conductors each carrying 2A pass perpendicular to the plane of the triangle through the corners B and C. The sides of the trianlge are AB=3cm, AC=4cm.
Answer» Correct Answer - `16.6 muT`

Discussion

87.	` H^(+), He^(+) and O^(++)` all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity . The masses of ` H^(+), He^(+) and O^(2+)` are `1 amu, 4 amu and 16 amu` respectively . ThenA. `H^(+)` ions will be deflected mostB. `O^(++)` ions will be deflected leastC. `He^(+)` and `O^(++)` ions will suffer same deflectionD. All ions will suffer the same deflection
Answer» Correct Answer - A::C

Discussion

88.	Why do magnetic lines of force prefer to pass through ferromagnetic materials?
Answer» This is because permeability `(mu)` and susceptibity `(chi_m)` of such materials are very high.

Discussion

89.	A conductor carrying current I is of the type as shown in figure. Find the magnetic field induction at the common centre O of all the three arcs.
Answer» Correct Answer - `(5mu_0Itheta)/(24pir)` We know that magnetic field induction at the centre of an arc subtending an angle `theta` is `B=(mu_0)/(4pi)I/rtheta` `:.` Total magnetic field induction at O due to current through the three arcs is `B=(mu_0)/(4pi)I[1/r-(1)/(2r)+(1)/(3r)]theta=(5mu_0Itheta)/(24pir)`

Discussion

90.	Deduce the expression for the magnetic field induction at the centre of a circular electron orbit of radius r, and angular velocity of orbiting electron `omega`.
Answer» The electron moving on a circular orbit acts like a current loop. Magnetic field induction in magnitude, at the centre of the current loop is `B=(mu_0)/(4pi i)(2pir)/(r)=(mu_02pi)/(4pir)(e/T)=(mu_0)/(2r)((e)/(2pi//omega))=(mu_0eomega)/(4pir)`

Discussion

91.	A particle of charge ` +q` and mass `m` moving under the influence of a uniform electric field ` E hat(i)` and uniform magnetic field ` B hat(k)` follows a trajectory from ` P to Q` as shown in fig. The velocities at `P and Q` are ` v hat(i)` and ` - 2v hat(j)` . which of the following statement(s) is/are correct ? A. `E=3/4((mv^2)/(qa))`B. Rate of work done by the electric field at P is `3/4[(mv^3)/(a)]`C. Rate of work done by the electric field at `P=0`D. Rate of work done by both fields at Q is zero
Answer» Correct Answer - A::B::D Increase in K.E=work done by electric field `1/2m(2v)^2-1/2mv^2=qE(2a)` or `E=3/4((mv^2)/(qa))` (b) Rate of work done by electric field at P =Electric force x Velocity at P `qExxv=q[3/4(mv^2)/(qa)]v`, `=3/4(mv^3)/(a)` (d) At Q, force acting on particle due to electric field is zero (as `2vhatj`) is perpendicular to both `Ehati` and `Bhatk`) and hence rate of work done by both fields is zero.

Discussion

92.	` H^(+), He^(+) and O^(++)` all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity . The masses of ` H^(+), He^(+) and O^(2+)` are `1 amu, 4 amu and 16 amu` respectively . ThenA. `H^+` will be deflected mostB. `O^(2+)` will be deflected mostC. `He^+` and `O^(2+)` will be deflected equallyD. all will be deflected equally
Answer» Correct Answer - A::C (c) As `r=(mv)/(qB)` and `K=1/2mv^2`, `r=(sqrt(2mK))/(qB)` i.e., `rpropsqrtm/q` [for same K and B] `r_(H+)propsqrt1/1` and `r_(O^(2+))` are equal, So `He^+` and `O^(2+)` will be deflected equally. (a) Since `(r_(H^+))/(r_(He^+))=(sqrt1//1)/(sqrt4//1)=1/2`, `r_(H^+)=1/2He^+=1/2r_(O^(2+))` `H^+` will be deflected most.

Discussion

93.	As shown in figure a cell is connected across two points A and B of a uniform circular conductor of radius r. Prove that the magnetic field induction at its centre O will be zero.
Answer» Let `l_1` and `l_2` be the lengths of the two circular arcs ACB and ADB respectively. `rho` be the resistance per unit length of the circular conductor. Resistance of circular arc `ACB, R_1=l_1rho` Resistance of circular arc `ADB, R_2=l_2rho` Let `I_1`, `I_2` be the current in arms ACB and ADB respectively. As these two arcs are connected in parallel, so the potential difference across them must be equal i.e., `I_1R_1=I_2R_2` or `I_1l_1rho=I_2l_2rho` or `I_1l_1=I_2l_2` ...(i) Magnetic field induction at O due to currrent in arc ADB is, `B_2=(mu_0)/(4pi)I_2/rtheta_2=(mu_0)/(4pi)(I_2l_2)/(r^2)` It is acting downwards, perpendicular to the plane of circular conductor. As `I_1l_1=I_2l_2`, so `B_1=B_2` and `vecB_1=-vecB_2` `:.` Resultant magnetic field at `O=vecB_1+vecB_2` `=(-vecB_1)+vecB_2=0`

Discussion

94.	An electron emmited by a heated cathode and accelerated through a potential difference of `20kV` enters a region with a uniform magnetic field of `015T`. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity (b) makes an angle of `30^@` with the initial velocity.
Answer» Here, `V=20kV=20xx10^3V` , `B=015T`. Let e, m be the charge and mass of the electron. If v is the velocity gained by electron when accelerated under a potential difference V, then `eV=1/2mv^2` or `v=((2eV)/(m))^(1//2)=((2xx16xx10^(-19)xx20xx10^3)/(90xx10^(-31)))^(1//2)=(8xx10^7)/(3)m//s` (a) Force on electron due to transverse magnetic field is `=Bev`, which is perpendicular to `vecB` as well as `vecv`. It will provide the required centripetal force to the electron for its circular motion. Therefore, the trajectory of electron in magnetic field is circular. Its radius r can be given by `Bev=(mv^2)/(r)` or `r=(mv)/(Be)=(9xx10^(-31)xx(8xx10^7//3))/(015xx16xx10^(-19))=10^-3m=1mm` (b) When electron makes an angle `theta` with the direction of magnetic field, its component velocity perpendicular to the field, `v_1=v sin 30^@=8/3xx10^7xx1/2=4/3xx10^7m//s`. Due to this velocity, the force acting on the electron due to magnetic field will be providing the required centripetal force, hence the electron will describe a circular path of radius `r_1`. The component velocity along the magnetic field `=vcos30^@=8/3xx10^7xxsqrt3/2=(4sqrt3)/(3)xx10^7m//s`. Due to this velocity, no force acts on the electron in magnetic field, and electron moves without change in its speed along the magnetic field. Therefore, due to two perpendicular component velocities, the electron describes a helical path. The radius of the helical path, `r=(mv_1)/(Be)=(9xx10^(-31)xx(4//3)xx10^7)/(015xx16xx10^(-19))=05xx10^-3m=05mm`

Discussion

95.	The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will beA. `(pi^(2))/(4sqrt(2))`B. `(pi^(2))/(8sqrt(2))`C. `(pi)/(2sqrt(2))`D. `(pi)/(4sqrt(2))`
Answer» Correct Answer - B

Discussion

96.	A compass needle, pivoted about the horizontal axis and forced to move in the magnetic meridian is observed to point along (i) vertical direction at place A, (ii) horizontal direction at place B. What are the angles of dip at the two places?
Answer» At place A, `delta=90^@` and at place B, `delta=0^@`

Discussion

97.	Two identical magnets with a length `10cm` and weights `50gf` each are arragned freely with their like poles facing in a vertical glass tube. The upper magnet hangs in air above the lower one so that the distance between the nearest poles of the magnets is `3mm`. Determine the pole strength of the poles of these magnets.
Answer» Correct Answer - `664Am` Let `m_1=m_2=m`, `r=3mm=3xx10^-3m` For hanging in air in balanced position, `F=50g f=50xx10^-3xx98N` As `F=(mu_0)/(4pi)(m_1m_2)/(r^2)` `:. 50xx10^-3xx98=10^-7xx(m.m)/((3xx10^-3)^2)` `m^2=9xx5xx98xx10^-1`, `m=6*64Am`.

Discussion

98.	In the figure are shown two long parallel current carrying wires I and II. Find the magnitudes and directions of the magnetic field induction at the point P, Q and R in the plane of paper.
Answer» Correct Answer - `2xx10^-5T_\|__r` to the plane of paper upwards; `10^-4T_\|__r` to the plane of paper downwards; `47xx10^-5T_\|__r` to the plane of paper upwards Resultant magnetic field induction at P is `B=B_1-B_2` `=(mu_0)/(4pi)(2I_1)/(r_1)-(mu_0)/(4pi)(2I_2)/(r_2)=(mu_0)/(4pi)xx2[I_1/r_1-I_2/r_2]` Here, `I_1=20A, r_1=01m`, `I_2=30A`, `r_2=030m` `B=10^-7xx2[(20)/(01)-(30)/(03)]=2xx10^-5T` It will be acting perpendicular to the plane of paper upwards. Resultant magntic field induction at Q is `B=(mu_0)/(4pi)xx2[(20)/(01)+(30)/(01)]` `=10^-7xx2xx500=10^-4T` It will be acting perpendicular to the plane of the paper downwards. Resultant magntic field induction at R is `B=(mu_0)/(4pi)xx2[(30)/(01)-(20)/(03)]=47xx10^-5T` It will be acting perpendicular to the plane of the paper upwards.

Discussion

99.	There are two parallel current carrying wires X and Y as shown in figure. Find the magnitude and direction of the magnetic field at points, P, Q and R.
Answer» Here, `I_1=10, I_2=15A, r=5cm=5xx10^-2m` Let `hatn` be the unit vector perpendicular to the plane of paper unwards. The magnetic field at P due to current in wire X is `vecB_1=(mu_0)/(4pi)(2I_1)/(r)hatn=10^-7(2xx10)/(5xx10^-2)hatn=4xx10^-5hatnT` The magnetic field at P due to current in wire Y is `vecB_2=(mu_0)/(4pi)(2I_2)/((r+r+r))(-hatn)=10^-7=(2xx15)/(15xx10^-2)(-hatn)=-2xx10^-5T` The resultant magnetic field at P is `vecB_P=vecB_1+vecB_2=(4xx10^-5)hatn-(2xx10^-5)hatn=2xx10^-5Thatn` `=2xx10^-5T` acting normally outwards. At point Q, both `vecB_1` and `vecB_2` will be acting normally inwards. So `vecB_Q=vecB_1+vecB_2=(mu_0)/(4pi)(2xx10)/(5xx10^-2)(-hatn)+(mu_0)/(4pi)(2xx15)/(5xx10^-2)(-hatn)` `=[10^-7xx(20)/(5xx10^-2)+10^-7xx(30)/(5xx10^-2)](-hatn)=(4xx10^-5+6xx10^-5)(-hatn)` `=10^-4T` acting normally inwards. At point R, `vecB_1` will be acting normally inwards and `vecB_2` will be acting normally outwards. So, `vecB_R=vecB_1+vecB_2=10^-7xx(2xx10)/((5+5+5)xx10^-2)(-hatn)+10^-7xx(2xx20)/((5xx10^-2))(hatn)` `=(-4/3xx10^-5+8xx10^-5)(hatn)=20/3xx10^-5T`, acting normally outwards

Discussion

100.	In the circuit, shown in figure. `R=5000Omega`. If key `K_1` is closed, galvanometer shows a deflection of 30 scale division. On closing key `K_2` and making `S=20Omega`, the deflection of galvanometer reduces to 15 division. The resistance of galvanometer is A. `50Omega`B. `30Omega`C. `20Omega`D. `15Omega`
Answer» Correct Answer - C `G=(RS)/(R-S)=(5000xx20)/(5000-20)=20Omega`

Discussion

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