InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A current carrying closed loop in the from of a right angle isoseles triangle `ABC` is placed in a unifrom magnetic fild acting along `AB`. If the magnetic force on the arm `BC` is `F`, the force on the arm `AC` is A. `-F`B. `F`C. `sqrt(2) F`D. `-sqrt(2) F` |
|
Answer» Correct Answer - A When a current carrying closed loop is placed in a unifrom magnetic field, it experiences no net force. `vec(F)_(AB) + vec(F)_(BC) + vec(F)_(CA) = 0`. But `vec(F)_(AB) = 0` as the side `AB` is parallel to magnetic field hence it wil experience no force due to magnetic field. Hence `vec(F)_(BC) + vec(F)_(CA) = 0 implies vec(F)_(BC) = -vec(F)_(CA)` |
|
| 102. |
A ring of radius R having unifromly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is `T_0.` Now a vertical magnetic field is switched on and ring is rotated at constant angular velocity `omega`. Find the maximum `omega` with which the ring can be rotated if the strings can withstand a maximum tension of `3T_0 //2.` |
|
Answer» In equilibrium, `2T_0=mg` or `T_0=(mg)/2….(i)` magnetic moment, `M=iA=(omega/(2pi)Q)(piR^2)` `tau=MB sin90^@=(omegaBQR^2)/2` Let `T_1 and T_2` be the tensions in the two strings when magnetic field is switched on `(T_1gtT_2)` For translation equilibrium, `T_1+T_2=mg....(ii)` For rotational equilibrium `(T_1-T_2)D/2=tau=(omegaBQR^2)/2` or `T_1-T_2=(omegaBQR^2)/D....(iii)` Solving Eqs. (ii) and (iii) we have `T_1=(mg)/2+(omegaBQR^2)/(2D)` As `T_1gtT_2` and maximum values of `T_1` can be `3T_0//2`, we have `(3T_0)/2=T_0+(omega_(max)BQR^2)/(2D) ((mg)/2=T_0)` `:. omega_(max)=(DT_0)/(BQR^2)` |
|
| 103. |
A proton and an `alpha-`particle enter a uniform magnetic field moving with the same speed. If the proton takes `25 mu s` to make 5 revolutions, then the periodic time for the `alpha-` particle would beA. `50 mu s`B. `25 mu s`C. `10 mu s`D. `5 mu s` |
|
Answer» Correct Answer - c `T=(2pim)/(Bq) or Tpropm/q` `T_(alpha)/(T_p)=(4m)/(2q)xxq/m=2` or `T_(alpha)=2[25/5]mus=10mus` |
|
| 104. |
A current carrying loop is free to turn in a uniform magnetic field. The loop will then come into equilibrium when its plane is inclined atA. `0^(@)` to the direction of the fieldB. `45^(@)` to the directin of the fieldC. `90^(@)` to the direction of the fieldD. `135^(@)` to the direction of the field |
| Answer» Correct Answer - a | |
| 105. |
A ring of radius `R`, made of an insulating material carries a charge `Q` uniformly distributed on it. If the ring rotates about the axis passing through its centre and normal to plane of the ring with constant angular speed `omega`, then the magnitude moment of the ring isA. `QomegaR^(2)`B. `(1)/(2)QomegaR^(2)`C. `Qomega^(2)R`D. `(1)/(2)Qomega^(2)R` |
|
Answer» Correct Answer - B `M=iA=ixxpiR^(2)` also `i=(Qomega)/(2pi)implies(1)/(2)QomegaR^(2)` |
|
| 106. |
A circular loop of area `0.01 m^(2)` carrying a current of `10 A`, is held perpendicular to a magnetic field of intensity `0.1 T`. The torque acting on the loop isA. ZeroB. `0.01 N-m`C. `0.001 N-m`D. `0.8 N-m` |
|
Answer» Correct Answer - A `tau=NiABsintheta=0` (`:. Theta=0^(@)`) |
|
| 107. |
In a hydrogen atom , the electron moves in an orbit of radius ` 0.5 A` making `10^( 16 ) revolutions per second . The magnetic moment associtated with the orbital motion of the electron is …… |
|
Answer» `i=q/t=("ne")/t=(10^16)/1xx1.6xx10^-19=1.6xx10^-3A` `M=ixxA=ipir^2=1.6xx10^-3xx3.14xx(0.5xx10^-10)^2` `=1.25xx10^-23Am^2` |
|
| 108. |
the intensity of the magnetic induction field at the centre of a single turn circular coil of radius 5 cm carrying current of 0.9 A isA. `36pixx10^(-7)T`B. `9pixx10^(-7) T`C. `36pixx10^(-6)T`D. `9pixx10^(-6) T` |
|
Answer» Correct Answer - a The intensity of magnetic induction filled `B=(mu_(0)i)/(2r) =(4pixx10^(-7)xx0.9)/(2xx5xx10^(-2))` `B=36 pi xx10^(-7) T` |
|
| 109. |
The magnetic moment of a current carrying loop is `2.1xx10^(-25) ampxxm^(2)`. The magnetic field at a point on its axis at a distance of `1 Å` isA. `4.2xx10^(-2)weber//m^(2)`B. `4.2xx10^(-3)weber//m^(2)`C. `4.2xx10^(-4)weber//m^(2)`D. `4.2xx10^(-5)weber//m^(2)` |
|
Answer» Correct Answer - A Field at a point `x` from the centre of a current carrying loop on the axis is `B=(mu_(0))/(4pi).(2M)/(x^(3))=(10^(-7)xx2xx2.1xx10^(-25))/(10^(-10))^(3)` `4.2xx10^(-32)xx10^(30)=4.2xx10^(-2) W//m^(2)` |
|
| 110. |
Current `i` is carried in a wire of length `L`. If the wire is turned into a circular coil, the maximum magnitude of torque in a given magnetic field `B` will beA. `(LiB^(2))/(2)`B. `(Li^(2)B)/(2)`C. `(L^(2)iB)/(4pi)`D. `(Li^(2)B)/(4pi)` |
|
Answer» Correct Answer - C `tau_(max)=NiAB=1xxixx(pir^(2))xxB` `(2pir=L,impliesr=(L)/(2pi))tau_(max)=pii((L)/(2pi))^(2)B=(L^(2)iB)/(4pi)` |
|
| 111. |
A particle of the charged `q` and `mass m` moves in a circular orbit of radius `r ` with angular speed ` omega` . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends onA. `omega and q`B. `omega,q and m`C. `q and m`D. `omega and m` |
|
Answer» Correct Answer - c The angular momentum L of the particle is given by `L=mr^2omega` `to` where `omega=2pin` `:.` Frequency `n=omega/(2pi)` Further `ip=qxxn=(omegaq)/(2pi)` Magnetic moment, `M=iA=(omegaA)/(2pi)xxpir^2` `:. M=(omegaqr^2)/2 implies M/L=(omegaqr^2)/(2mr^2omega)=q/(2m)`. |
|
| 112. |
The maximum current that can be measured by a galvanometer of resistance `40 Omega` is 10 mA . It is converted into a voltmeter that can read upto 50 V . The resistance to be connected in series with the galvanometer is ... (in ohm )A. 2010B. 4050C. 5040D. 4960 |
|
Answer» Correct Answer - d To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is `R=V/(I_(g))-G=50/(10xx10^(-3)) -40 =5000-40 =4960 Omega` |
|
| 113. |
A particle of the charged `q` and `mass m` moves in a circular orbit of radius `r ` with angular speed ` omega` . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends onA. `omega` and `q`B. `omega` `q`and `m`C. `q`and `m`D. `omega`and `m` |
|
Answer» Correct Answer - C The effective current `i=(qomega)/(2pi)` and `A=pir^(2)` Magnetic moment `M=iA-(1)/(2)qomegar^(2)` Angular moment `L=Iomega=mr^(2)omegaimplies(M)/(L)=(q)/(2m)` |
|
| 114. |
A uniform magnetic field `B` and a uniform electric field `E` act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated isA. C. D. |
|
Answer» Correct Answer - C For undeviated motion `|vecF_(e)|=|vecF_(m)|`, which happened when `vec(v),vec(E)` and `vec(B)` are mutually perpendicular to each other. |
|
| 115. |
The magnitude of magnetic moment of the current loop in the figure is A. `la^(2)`B. `sqrt(2) la^(2)`C. zeroD. none of the above |
| Answer» Correct Answer - b | |
| 116. |
A particle of charge q and mass `m` is moving along the x-axis with a velocity `v` and enters a region of electric field `E` and magnetic field `B` as shown in figures below. For which figure the net force on the charge may be zero?A. B. C. D. |
|
Answer» Correct Answer - B The charge will not experience any force if `|vecF_(e)|=|vecF_(m)|`. This condition is satisfied in option |
|
| 117. |
A uniform magnetic field `B=B_(0)hatj` exists in space. A particle of mass m and charge q is projected towards X-axis with speed v from a point (a,0, 0). The maximum value of v for which the particle does not hit the Y-Z plane isA. `(Bqa)/m`B. `(Bqa)/(2m)`C. `(Bq)/(am)`D. `(Bq)/(2am)` |
| Answer» Correct Answer - a | |
| 118. |
A particle of mass `m` and charge `q` travelling at a velocity `v` along `x`-axis at time `t=0`. It enters a region in which electric field, `E` is present along `z`-axis and magnetic field, `B` is present along `y`-axis. The distance covered upto time `t=0.2xx10^(-1) sec,` along `x`-axis is `v//50` (in meters). find the value of `v`: A. `(50 E)/B`B. `E/(50 B)`C. `E/(20B)`D. `E/B` |
|
Answer» Correct Answer - D Distance covered in time `t=v//50` (given) From, `x=vt=vxx0.02=v//50` Thus, acceleration of particle `=0` So, `F_("net")=0` `F_(e)=F_(m)` `qE=qvB` `v=(E )/(B)` |
|
| 119. |
A particle of mass `m` having negative charge `-q` is projected at an angle `theta` with `x`-axis. There exists uniform electric field `E` and a unifrom magnetic field `B` along `x`-axis. The particle will return to its intial point once if (`n` is some integer) A. `n=(Bvcostheta)/(piE)`B. `n=(2Bvcostheta)/(piE)`C. `n=(Bvsintheta)/(piE)`D. `n=(2Bvsintheta)/(piE)` |
|
Answer» Correct Answer - A Let particle completes `n` revolutions in time `t`, then `t=nT` `implies t=(n2pim)/(qB)` If particle has to return to its initial point, the displacement along `x`-axis during this time should be zero. `s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)` ` implies 0=vcosthetat-(1)/(2)(qE)/(m)t^(2)` ` impliesvcostheta=(qE)/(2m)t` ` impliesvcostheta=(qE)/(2m)(n2pim)/(qB)impliesn=(Bvcostheta)/(piE)` |
|
| 120. |
In mass spectrometer used for measuring the masses of ions, the ions are initaily accerlerated by an electric potential `V` and then made to describe semicircular paths of radius `R` using a magnetic field `B`.if `V` and `B` are kept constant, the ratio `(("charg e on the ion")/("mass of the ion"))` will be propertional to:A. `(1)/(R)`B. `(1)/(R^(2))`C. `R^(2)`D. `R` |
|
Answer» Correct Answer - B (b) If an ion of charge `q` mass `m` moves in magnetic fied of flux density `B` with velocity `v`, its path is circular. The magnetic force `qvB` acting on the ion provides required centripeteal force `(mv^(2))/(R)` on the ion moving in a circular path of radius `R`. `(mv^(2))/(R) = qvB implies R = (mv)/(qB)` The angular frequency of rotation of the ions about the vertical field `B` is given by `omega = (v)/(R) = (qB)/(m)` Energy of ion is given by `E = (1)/(2) mv^(2) = (1)/(2) m(R omega)^(2)` `implies E = (1)/(2) mR^(2) B^(2) (q^(2))/(m^(2))` or `E = (1)/(2) (R^(2) B^(2) q^(2))/(m)` ....(i) If ions are accelearated by electric potential `V`, then energy attained by ions, `E = qV` From Eqs.(I) and (ii), we get `qV = (1)/(2) (R^(2) B^(2) q^(2))/(m)` or `(q)/(m) = (2V)/(R^(2) B^(2))` If `V` and `B` are kept constant , then `(q)/(m) prop (1)/(R^(2))` |
|
| 121. |
Calculate the force on a current carrying wire in a uniform magnetic field as shown in Fig. 1.59. |
|
Answer» The net force from `A` to `B` is `dvecF=I(dvecLxxvecB)` `int_A^BdvecF=int_A^PI[dvecL_1xxvecB]+int_P^QI[dvecL_2xxvecB]+int_Q^RI[dvecL_3xxvecB]+int_R^TI[dvecL_4xxvecB]+int_T^BI[dvecL_5xxvecB]` The entire path can be broken down into element vectors joined to each other in sequence. We known, from polygon law of addition of vectors, that vector joining the tail of the first vector to the head of the last vector is the resultant. `vecF=I(vecLxxvecB)` where `|vecL_1|=a+sqrt(c^2-b^2+2r+d` `F_("net") =IB(a+sqrt(c^2-b^2)+2r+d)` and its direction is upward on plane of paper. |
|
| 122. |
A wire of length `L` is shaped into a circle and then folded as shown. Magnetic moment of the frame when a current `i` is passed through it is: A. `(L^(2)i)/(4pi)`B. `(L^(2)i)/(4sqrt2pi)`C. `(L^(2)i)/(4)`D. `(L^(2)i)/(8pi)` |
|
Answer» Correct Answer - B Since `L=2pir, r=(L)/(2pi)` Area of circle `pir^(2)=(piL^(2))/(4pi^(2))=(L^(2))/(4pi)` Area of each half of circle `=(L^(2))/(8pi)` Magnetic moment of each half `=(L^(2)i)/(8pi)` The two magnetic moments are at right angles to each other. Magnetic moment of the frame `=sqrt(((L^(2)i)/(8pi))^(2)+((L^(2)i)/(8pi))^(2))=(sqrt2.L^(2)i)/(8pi)=(L^(2)i)/(4sqrt(2pi))` |
|
| 123. |
A constant current `I` is flowing through a circular coil placed in uniform magnetic field `vecB` as shown. Then: A. The loop is in stable equilibriumB. The loop is in unstable equilibriumC. The torque acting on the loop is maximumD. The torque acting on the loop is `(1)/(sqrt(2))` times the maximum torque |
|
Answer» Correct Answer - C Torque acting on the loop is `tau=MxxB=Mbsintheta` Where `theta=` angle between `M` and `B` In the figure shown `theta=90^(@)` because `B` is in place of loop. Or `tau=MB=`maximum torque Hence, (c) is correct and choice (d) is wrong. Since, torque is acting on the loop so loop is not in equilibrium. Hence, choices (a) and (b) are wrong. |
|
| 124. |
The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction `1.5 xx 10^-2 T.` A current of 10.0 ampere flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the lines of induction, then the torque acting on the loop isA. 6000 newton `xx` metreB. zero newton `xx` metreC. `1.2xx10^(-2)` newton `xx` metreD. `6xx10^(-4)` newton `xx` metre |
|
Answer» Correct Answer - D Torque `tau` acting on a current carrying coil of area `A` placed in a magnitude field of induction `B` is given by: `tau=NIBAsintheta` Where `I=` current in the coil, `theta=` angle which the normal to the plane of the coil makes with the lines of induction `B`. Here `N=1,B=1.5xx10^(-2)tesla,` `A=0.05xx0.08=40xx10^(-4)m^(2),` `I=10.0amp,=90^(@)=pi//2` `tau=(1.5xx10^(-2))(10.0)xx(1)(40xx10^(-4))sin((pi)/(2))` `=6xx10^(-4)` newton `xx` meter |
|
| 125. |
A current I flows in a circular coil of radius r. If the coil is placed in a uniform magnetic field B with its plane parallel to the field, magnitude of the torque that acts on the coil isA. ZeroB. `2piriB`C. `pir^(2)iB`D. `2pir^(2)iB` |
|
Answer» Correct Answer - C |
|
| 126. |
A magnetic field of `(4.0xx10^-3hatk)T` exerts a force `(4.0hati+3.0hatj)xx10^-10N` on a particle having a charge `10^-9C` and moving in te `x-y` plane. Find the velocity of the particle.A. `-75 hati+100 hatj`B. `-100 hati +75 hatj`C. `25 hati+2hatj`D. `2hati+25 hatj` |
|
Answer» Correct Answer - a From Lorentz force, F=q(VxB) Given, `F=(4hati+3hatj) xx10^(-10)N, q=10^(-9)C` `B=4xx10^(-3) hatk T` `:. (4hati+3hatj) xx10^(-10) =10^(-9) (ahati+bhatj) xx(4xx10^(-3)) hatk` Solving, we get `a=-75, b=100 implies v =-75 hat i+100 hatj` |
|
| 127. |
A circular current carrying loop of `100` turns and radius `10 cm` is placed on `X-Y` plane. A unifromn magnetic field `B = (-hat(i)+hat(k)) T` is present in the region. If current in the loop is `5 A`, then match the Couloumb 1 with Couloum II and select the correct option from the the given codes. A. `i-p,ii--q,r,iii-s`,B. `i-q,r,ii-q,s,iii-p`C. `i-s,q,ii-p,iii-r`,D. `i-r,ii-p,iii-q,s`, |
|
Answer» Correct Answer - B (i) As, the center in `XY` plane is anti-clockwise, so moment will be along `Z`-axis by right hand thumb rule. (ii) `tau = M xx B` `|tau| = |M| |B| sin theta = (5xx pi xx (0.1)^(2) xx 100 sqrt(2))/(sqrt(2))` (here, `theta = 45^(@)`) (iii) Net force on a closed loop carrying current in a unifrom magnetic feild is zero. Hence, `(i) rarr (q,r), (ii) rarr (q,x), (iii) rarr (p)` |
|
| 128. |
The magnetic field at O due to current in the infinite wire forming a loop as a shown in Fig. A. `(mu_(0)I)/(4pid)(cosphi_(1)+cosphi_(2))`B. `(mu_(0))/(4pi)(2I)/d(tan theta_(1)+tan theta_(2))`C. `(mu_(0))/(4pi)(I)/d(sin phi_(1)+sin phi_(2))`D. `(mu_(0))/(4pi)(I)/d(cos theta_(1)+cos theta_(2))` |
|
Answer» Correct Answer - B `B=(mu_(0)I)/(4pid) (sin theta_(1)+sin theta_(2))` But `theta_(1)+phi_(1)=90^(@)` or `theta_(1)=90^(@)-phi` `sin theta_(1)=sin(90^(@)-phi_(1))=cos phi_(1)` Similarly, `sin theta_(2)=cos phi_(2)` `B=(mu_(0)I)/(4pid) (cos phi_(1)+cos phi_(2))` |
|
| 129. |
Equal currents `i=1` A are flowing through the wires parallel to y-axis located at `x=+1m, x=+2m, x=+4m` and so on...., etc. but in opposite directions as shown in Fig The magnetic field (in tesla) at origin would be A. `-1.33xx10^(7) hatk`B. `1.33xx10^(-7) hatk`C. `2.67xx10^(-7) hatk`D. `-2.67xx10^(-7) hatk` |
|
Answer» Correct Answer - B `vec(B)=(mu_(0))/(2pi)[(1/1+1/4+1/16+....) hatk-(1/2+1/8+....)hatk)]` `=(mu_(0))/(2pi)[(1/(1-1/2))hatk-1/2(1/(1-1/4))hatk]` `2/3.(mu_(0))/(2pi) hatk` `=(2/3xx2xx10^(-7))hatK` `=1.33xx10^(-7) hatK` |
|
| 130. |
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. `4 eV`B. `2 eV`C. `8 eV`D. `6 eV` |
|
Answer» Correct Answer - C `r=sqrt(2mK)/(qB)impliesqpropsqrtmKimpliesKprop(q^(2))/(m)` `(K_(alpha))/(K_(p))=(q_(alpha)/(p_(p)))^(2)xx(m_(p))/(m_(alpha))implies(k_(alpha))/(8)=((2q_(p))/(q_(p)))^(2)xx(m_(p))/(4m_(p))=1` ` implies K_(alpha)=8eV` |
|
| 131. |
Mixed `He^(+)` and `O^(2+)` ions (mass of `He^(+)`=4 amu and that of `O^(2+)=16` amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same thenA. `He^(+)` ions will be deflected more than those of `O^(2+)`B. `He^(+)` ions will be be deflected less than those of `O^(2+)`C. All the ions will be deflected equallyD. No ions will be deflected |
|
Answer» Correct Answer - C `r=sqrt(2mK)/(qB)implies rpropsqrt(m)/(q)implies(r_(He^(+)))/(r_(o^(++)))=sqrt(m_(He^(+))/(m_(O^(++))))xx(q_(O^(++)))/(q_(He^(++)))` `sqrt((4)/(16))xx(2)/(1)=(1)/(1)`. Then will deflect equally. |
|
| 132. |
A particle of mass `0.6g` and having charge of `25 n_(C)` is moving horizontally with a uniform velocity `1.2xx10^(4)ms^(-1)` in a uniform magnetic field, then the value of the magnetic induction is `(g=10ms^(-2))`A. ZeroB. 10 TC. 20 TD. 200 T |
|
Answer» Correct Answer - C |
|
| 133. |
An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion wil beA. Straight line along the x-directionB. A circle in the xz-planeC. A circle in the yz-planeD. A circle in the xy-plane |
|
Answer» Correct Answer - B |
|
| 134. |
A wire carrying current `I` is shaped as shown. Section `AB` is a quarter circle of radius `r`. The magnetic field is directed A. along the bisector of the anlge `ACB`, away from `AB`B. along the bisector of the angle `ACB`, towards `AB`C. perpendicular to the plane of the paper, directed into the paperD. at an angle `pi//4` in the plane of the paper |
| Answer» Correct Answer - C | |
| 135. |
A conductor (shown in the figure) carrying constant current `I` is kept in the `x-y` plane in a uniform magnetic field ` vec(B)`. If ` vec(F)` is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are) A. If `vecB` is along `hatz, Fprop (L+R)`B. If `vecB` is along `hatx, F=0`C. If `vecB` is along `haty, Fprop (L+R)`D. If `vecB` is along `hatz, F=0` |
|
Answer» Correct Answer - (a,b,c) As `vecB` is uniform `implies` Wire can be replaced by a straight current carrying conductor. `implies vecF=i(veclxxvecB)` `=(2(L+R)hatixxvecB)` if `vecB` is along x-axis `implies vecF=0` otherwise `Fprop(L+R)` |
|
| 136. |
A charged particle is moving in a uniform magnetic field and losses 4% of its KE. The radius of curvature of its path change byA. 0.02B. 0.04C. 0.1D. None of these |
|
Answer» Correct Answer - a `r=(mv)/(qB)` `KE=K=1/2 mv^(2)` `:. mv=sqrt(2Km)` `:. r=(mv)/(qB)=(sqrt(2Km))/(qB) ` or `r prop sqrt(K)` or `r=eK^(1/2)`, where e is a constant . or `(dr)/(dr) =(dK^(1/2))/(dr) =e/2 (ec)/(2sqrt(K)) ` or `(ec)/(2sqrt(K)) (dK)/(dr) -sqrt(K)` or `e(DeltaK)/(Deltar) =2sqrt(K) ` or `(Deltar)/r=(cDeltaK)/(2sqrt(Kc)sqrt(K)) =(DeltaK)/(2K)` or `(Deltar)/rxx100=(DeltaK)/(2K) xx100 =2% [ :. (DeltaK)/K xx100 =4%]` |
|
| 137. |
A charged particle of mass m and charge q, a uniform magneric field B acting into the plane. The plane is frictional having coefficient of friction `mu`. The speed of charged particle just before entering into the region is `V_(0)`. The radius of curvature of the path after the time `(v_(0))/(2mug)` isA. `(mv_(0))/(qB)`B. `(mv_(0))/(2qB)`C. `(mv_(0))/(4qB)`D. None of these |
|
Answer» Correct Answer - b Here, centripetal acceleration is provided by magnetic force but tangential acceleration is provided by force of friction `(mv^(2))/r=qvB` `:. r=(mv)/(qB)` Tangential acceleration `=a=(-mumg)/m` `v=-mug((v_(0))/(2mug)) =(v_(0))/2` `:. r=(mv)/(qB)=(mv_(0))/(2qB)` |
|
| 138. |
A charge particle is moving in a circular path in a magnetic field. A resistive force starts acting on the particle whose direction is oppositely directed to its motion and magnitude is directely proportional to its velocity. Now the particle starts moving in a spiral path, thenA. Angular velocity of the particle decreases continuously.B. Angular momentum of the particle remains constantC. Magnetic field at lying only perpendicular to the plane of coilD. Net force acting on the particle remains constant. |
|
Answer» Correct Answer - C Radius `r` of particle in spiral path at any time when velocity is `v` `r=(mv)/(qB) omega=v/r=(qB)/m. implies` constant Angular momentum `L=mvr=qBr^(2)` decreases as `r` decreases If particle is moving in a plane then its velocity and acceleration will be that plane. The force on the particle will also be in that plane. It means magnetic field should be perpendicular to that plane. Mangnetic force on the particle `=qvB` Resisitive on the particle `=kv`, where `k` is a constant As velocity decreases so net force acting on the particle will decreases. |
|
| 139. |
A moving coil galvanometer has `N` numbr of turns in a coil of effective area `A`, it carries a current `I`. The magnetic field `B` is radial. The torque acting on the coil isA. `NA^(2)B^(2)I`B. `NABI^(2)`C. `N^(2)ABI`D. NABI |
|
Answer» Correct Answer - D `tau=Mbsinthetaimpliestau_(max)=NiAB, (theta=90^(@))` |
|
| 140. |
Two wires of the same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment isA. `2:pi`B. `pi:2`C. `pi:4`D. `4:pi` |
|
Answer» Correct Answer - C |
|
| 141. |
In a certain region of space, there exists a uniform and constant electric field of strength E along x-axis and uniform constant magnetic field of induction B along z-axis. A charge particle having charge q and mass m is projected with speed v parallel to x-axis from a point (a, b, 0). When the particle reaches a point 2a, `b//2`, 0 its speed becomes 2v. Find the value of electric field strength in term of m, v and co-ordinates.A. `(3)/(2)(mv^(2))/(qa)`B. `(mv^(2))/(qB)`C. `(2mv^(2))/(qBa)`D. `(3)/(2)vB` |
|
Answer» Correct Answer - A Change in kinetic energy is only due to work done by electric field. Hence work done by electric field is change in` KE`: `qEa=(1)/(2)m[(2v)^(2)-v^(2)]impliesE=(3mv^(2))/(2qa)` |
|
| 142. |
Assertion: Magnetic field at a point on the surface of long cylinderical wire is maximum. Reason: For any other point, closed loop perpendicular to the wire and of radius equal to the distance between axis of wire and given point will enclose less current.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - C For any point outside the wire enclosed current will be same. |
|
| 143. |
A semicircular ring is present in the uniform magnetic field. Magnetic field is perpendicular to loop of ring. Assertion: Force `vec(F)` on each element of ring is different Reason: Net force on ring must be perpendicular to magnetic field.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - B Magnetic force is always perpendicular to magnetic field and small element. |
|
| 144. |
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-y plane. If the current in the loop is i, the resultant magnetic field due to two semicircular parts at their common centre isA. `(mu_(0)i)/(2sqrt(2)R)`B. `(mu_(0)i)/(2R)`C. `(mu_(0)i)/(4R)`D. `(mu_(0)i)/(sqrt(2)R)` |
|
Answer» Correct Answer - A Due to semi-circular loop in `x-y` plane `B_(1)=(mu_(0)i)/(4R)` along `Z`-axis. Due semi-circular loop in `x-z` plane `B_(2)=(mu_(0)i)/(4R)` along `Y`-axis `B_(1)` is `bot^(ar)` to `B_(2)` `B_(R)=sqrt(B_(1)^(2)+B_(2)^(2))=sqrt(2). (mu_(0)i)/(4R)=(mu_(0)i)/(2sqrt(2)R)` |
|
| 145. |
A square coil of side 10 cm has 20 turns and carries a current of 12 A. the coil is suspended vartically and the normal to the plane of the coil, makes an angle `theta` with the direction of a uniform horizontal magnetic field of 0.80 T. if the torque, experienced by the coil, equals 0.N-m , the value of ` theta ` isA. `0^(@)`B. `pi/2 ` radC. `pi/3` radD. `pi/6` rad |
|
Answer» Correct Answer - d Area of coil A =`side^(2) =(0.1)^(2) =0.01 m^(2)`, Number of turns, N=20 current I=12 A Normal to the coil make an angle `theta` with the direction of B, magnetic field , B=0.80 T. torque, experienced by the coil , `tau=0.96 N-m` since, total torque on the coil , `tau=(NIA) B sin theta` Substituting the values in above formula, we get 0.96 N-m =20x12 A x `0.01 m^(2) xx0.80 T xx sin theta` `sin theta=0.96/1.92 =1/2` `theta=(pi)/6` rad |
|
| 146. |
A small coil of `N` turns has an effective area `A` and carries a current `I`. It is suspended in a horizontal magnetic field `vecB` such that its plane is perpendicular to `vecB`. The work done in rotating it by `180^(@)` about the vertical axis isA. `NAIB`B. `2NAIB`C. `2piNAIB`D. `4piNAIB` |
|
Answer» Correct Answer - B `W=MB(costheta_(1)-costheta_(2))` ` =(NiA)(Bcos 0^(@)-cos 180^(@))=2NAIB` |
|
| 147. |
An electron is moving in a circular path under the influence fo a transerve magnetic field of `3.57xx10^(-2)T`. If the value of `e//m` is `1.76xx10^(141) C//kg`. The frequency of revolution of the electron isA. `62.8 MHz`B. `6.28 MHz`C. `1 GHz`D. `100 MHz` |
|
Answer» Correct Answer - C `f = (eB)/(2pi m)` `f = (1.76xx10^(11)xx3.57xx10^(-2))/(2xx3.14)` `f = 10^(9) Hz` or `1 GHz` |
|
| 148. |
A 250-turns recantagular coil of length 2.1 cm and width 1.25 cm carries a current of `85 muA` and subjected to magnetic field of strength `0.85 T`. Work done for rotating the coil by `180^(@)` against the torque isA. `4.55 muJ`B. `2.3 muJ`C. `1.5 mu J`D. `9.1 mu J` |
|
Answer» Correct Answer - D Work `= MB [cos theta_(1) - cos theta_(2)]` Work `= MB [cos 0 - cos 180^(@)]` `W = NiAB [1 - (-1)]` `W = 9.1 muJ` |
|
| 149. |
Which of the following statements is correct?A. If the moving charged particle enters into a region of magnetic field form outside, it does not complete a circular path.B. If a moving charged particle traces a helical path in a uniform magnetic field, the axis of the helix is parallel to the magnetic field.C. The power associated with the force exerted by a magnetic field on a moving charged particle moving charged particle is always equal to zero.D. If in a region a uniform magnetic field and a uniform electric field both exist, a charged particle moving in this region cannot trace a circular path. |
|
Answer» Correct Answer - (a,b,c,d) Option (a) and (b) are theoretically facts. As in case of moving charged particle in magnetic field `vecF_(mag)botvecv`, hence power ass- ociated will be zero (option (c) is correct ). If both the elctric and mangnetic fields exist: If `vecB||vecE`, the path of the charged particle will be helical. If `vecB` is not parallel to `vecE`, the radius of the charged particle will not be constant. Hence, the path will not be circular (option(d) is correct). |
|
| 150. |
Figure shows an equilateral triangle ABC of side l carrying currents as shown, and placed in a uniform magnetic field B perpendicular to the plane of triangle. The magnitude of magnetic force on the triangle is A. ilBB. 2ilBC. 3ilBD. Zero |
|
Answer» Correct Answer - A |
|