1
PAST YEARS AIPMT QUESTIONS CATEGORY : MEDICAL
PHYSICS
Moving Charges and Magnetism 1.
2.
3.
4.
5.
6.
7.
A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes [1988] (a) inclined at 45° to the magnetic field (b) inclined at any arbitrary angle to the magnetic field (c) parallel to the magnetic field (d) perpendicular to the magnetic field Tesla is the unit of [1988] (a) magnetic flux (b) magnetic field (c) magnetic induction (d) magnetic moment Energy in a current carrying coil is stored in the form of [1989] (a) electric field (b) magnetic field (c) dielectric strength (d) heat The total charge induced in a conducting loop when it is moved in a magnetic field depends on (a) the rate of change of magnetic flux [1990] (b) initial magnetic flux only (c) the total change in magnetic flux (d) final magnetic flux only The magnetic induction at a point P which is at a distance of 4 cm from a long current carrying wire is 10–3 T. The field of induction at a distance 12 cm from the current will be [1990] (a) 3.33 × 10–4 T (b) 1.11 × 10–4 T (c) 3 × 10–3 T (d) 9 × 10–3 T A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is [1991] (a) 25 keV (b) 50 keV (c) 200 keV (d) 100 keV A uniform magnetic field acts at right angles to
8.
9.
10.
11.
12.
the direction of motion of electron. As a result, the electron moves in a circular path of radius 2cm. If the speed of electron is doubled, then the radius of the circular path will be [1991] (a) 2.0 cm (b) 0.5 cm (c) 4.0 cm (d) 1.0 cm The magnetic field at a distance r from a long wire carrying current i is 0.4 tesla. The magnetic field at a distance 2r is [1992] (a) 0.2 tesla (b) 0.8 tesla (c) 0.1 tesla (d) 1.6 tesla A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is [1992] (a) 2.4 N (b) 1.2 N (c) 3.0 N (d) 2.0 N To convert a galvanometer into an ammeter, one needs to connect a [1992] (a) low resistance in parallel (b) high resistance in parallel (c) low resistance in series (d) high resistance in series. A coil carrying electric current is placed in uniform magnetic field, then [1993] (a) torque is formed (b) e.m.f is induced (c) both (a) and (b) are correct (d) none of the above A charge moving with velocity v in X-direction is subjected to a field of magnetic induction in negative X-direction. As a result, the charge will (a) remain unaffected [1993] (b) start moving in a circular path Y–Z plane (c) retard along X-axis (d) move along a helical path around X-axis
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
2 13.
14.
15.
An electron enters a region where magnetic field (B) and electric field (E) are mutually perpendicular, then [1994] (a) it will always move in the direction of B (b) it will always move in the direction of E (c) it always possesses circular motion (d) it can go undeflected also A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying same current. The strength of magnetic field far away is [1995, 97, 99] (a) twice the earlier value (b) same as the earlier value (c) one-half of the earlier value (d) one-quarter of the earlier value At what distance from a long straight wire carrying a current of 12 A will the magnetic field
I C O
B
19.
be equal to 3 ´ 10 -6 Wb / m 2 ? [1995] 2 2 (a) 8 ´ 10 m (b) 12 ´ 10 m (c) 18 ´ 10 -2 m
16.
17.
18.
(d) 24 ´ 10 -2 m r The magnetic field ( dB ) due to a small element r (dl) at a distance (r ) and element carrying current i is [1996] r r r m0 æ d l ´ r ö i (a) dB = 4p çè r ÷ø r r m0 2 æ d l ´ rr ö = dB i (b) 4p çè r 2 ÷ø r r m 0 2 æ d l ´ rr ö = dB i (c) 4p çè r ÷ø r r m0 æ d l ´ rr ö (d) dB = 4p i ç 3 ÷ è r ø A 10 eV electron is circulating in a plane at right angles to a uniform field at magnetic induction 10– 4 Wb/m2 (= 1.0 gauss). The orbital radius of the electron is [1996] (a) 12 cm (b) 16 cm (c) 11 cm (d) 18 cm Two equal electric currents are flowing perpendicular to each other as shown in the figure. AB and CD are perpendicular to each other and symmetrically placed with respect to the current flow. Where do we expect the resultant magnetic field to be zero? [1996]
A
20.
21.
22.
23.
I
D
(a) on AB (b) on CD (c) on both AB and CD (d) on both OD and BO A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength 20 Vm–1 and 0.5 T respectively at right angles to the direction of motion of the electrons. Then the velocity of electrons must be [1996] (a) 8 m/s (b) 20 m/s (c) 40 m/s (d) 1 m / s 40 A galvanometer of resistance 20 W gives full scale deflection with a current of 0.004 A. To convert it into an ammeter of range 1 A, the required shunt resistance should be (a) 0.38W (b) 0.21W [1996] (c) 0.08W (d) 0.05W A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will (a) continue to move due east [1997] (b) move in a circular orbit with its speed unchanged (c) move in a circular orbit with its speed increased (d) gets deflected vertically upwards. Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. The force of attraction per unit length between the two wires is [1998] (a) 2 × 10–7 N/m (b) 2 × 10–8 N/m (c) 5 × 10–8 N/m (d) 10–7 N/m A galvanometer having a resistance of 8 ohms is shunted by a wire of resistance 2 ohms. If the total current is 1 amp, the part of it passing through the shunt will be [1998] (a) 0.25 amp (b) 0.8 amp (c) 0.2 amp (d) 0.5 amp
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
24.
25.
26.
27.
28.
A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic inductions at their centres will be [1998] (a) 2 : 1 (b) 1 : 4 (c) 4 : 1 (d) 1 : 2 Magnetic field intensity at the centre of a coil of 50 turns, radius 0.5 m and carrying a current of 2 A is [1999] (a) 0.5 × 10–5 T (b) 1.25 × 10–4 T (c) 3 × 10–5 T (d) 4 × 10–5 T When a proton is accelerated through 1 V, then its kinetic energy will be [1999] (a) 1840 eV (b) 13.6 eV (c) 1 eV (d) 0.54 eV If a long hollow copper pipe carries a current, then magnetic field is produced [1999] (a) inside the pipe only (b) outside the pipe only (c) both inside and outside the pipe (b) no where Two long parallel wires P and Q are both perpendicular to the plane of the paper with distance of 5 m between them. If P and Q carry currents of 2.5 amp and 5 amp respectively in the same direction, then the magnetic field at a point half-way between the wires is [2000] 3m 0 m0 (a) (b) 2p p m 3m 0 (d) 0 2p 2p A proton moving with a velocity 3 × 105 m/s enters a magnetic field of 0.3 tesla at an angle of 30º with the field. The radius of curvature of its path will be (e/m for proton = 108 C/kg) [2000] (a) 2 cm (b) 0.5 cm (c) 0.02 cm (d) 1.25 cm r In a certain region of space electric field E and r magnetic field B are perpendicular to each other and an electron enters in region perpendicular r r to the direction of B and E both and moves undeflected, then velocity of electron is [2001] r E r r (a) r (b) E ´ B B r B r r (c) r (d) E × B E
31.
(a)
30.
qB mp
(b)
qB 2 pm
qBE qB (d) 2pm 2 pE A galvanometer can be converted into a voltmeter by connecting [2002] (a) A high resistance in parallel (b) A low resistance in series (c) A high resistance in series (d) A low resistance in parallel A wire carries a current. Maintaining the same current it is bent first to form a circular plane coil of one turn which produces a magnetic field B at the centre of the coil. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre of the double loop, caused by the same current is [2002] (a) 4B (b) B/4 (c) B/2 (d) 2B A particle having charge q moves with a velocity r v through a region in which both an electric r r field E and a magnetic field B are present .The force on the particle is [2002] r r r r r r (a) qE + q( B ´ v ) (b) qE .( B ´ v ) r r r r r r (c) qv + q( E ´ B) (d) qE + q(v ´ B) A charged particle moves through a magnetic field in a direction perpendicular to it. Then the (a) velocity remains unchanged [2003] (b) speed of the particle remains unchanged (c) direction of the particle remains unchanged (d) acceleration remains unchanged A long solenoid carrying a current produces a magnetic field B along its axis. If the current is double and the number of turns per cm is halved, the new value of the magnetic field is [2003] (a) 4B (b) B/2 (c) B (d) 2B A galvanometer acting as a voltmeter will have [2004] (a) a low resistance in series with its coil. (b) a high resistance in parallel with its coil (c) a high resistance in series with its coil (d) a low resistance in parallel with its coil
(c)
32.
33.
34.
(c)
29.
3 A charged particle of charge q and mass m enters r perpendicularly in a magnetic field B . Kinetic energy of the particle is E; then frequency of rotation is [2001]
35.
36.
37.
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
4 38.
39.
A galvanometer of 50 ohm resistance has 25 divisions. A current of 4 × 10–4 ampere gives a deflection of one per division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of [2004] (a) 2450 W in series (b) 2500 W in series. (c) 245 W in series. (d) 2550 W in series. An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to [2005] v B v B (b) (c) (d) B B v v A very long straight wire carries a current I. At the instant when a charge + Q at point P r has velocity v , as shown, the force on the charge is [2005] Y
43.
44.
(a) 40.
45.
41.
X
(a) along OY (b) opposite to OY (c) along OX (d) opposite to OX When a charged particle moving with velocity r v is subjected to a magnetic field of induc-
æ charge on the ion ö ç ÷ will be è mass of the ion ø proportional to [2007] (a) 1/R2 (b) R2 (c) R (d) 1/R A charged paritcle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment µ is given by [2007] (a) qvR2 (b) qvR2/2 (c) qvR (d) qvR/2 Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion [2007] (a) depends on both R and v (b) is independent of both R and v (c) depends on R and not on v (d) depends on v and not on R Q
the ratio
46. O
A beam of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move (a) in a circular orbit [2007] (b) along a parabolic path (c) along a straight line (d) in an elliptical orbit. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular path of radius R using a magnetic field B. If V and B are kept constant,
47.
P
uur tion B , the force on it is non-zero. This implies
that
[2006]
F1
ur uur (a) angle between v and B can have any
42.
value other than 90° ur uur (b) angle between v and B can have any value other than zero and 180° ur uur (c) angle between v and B is either zero or 180° ur uur (d) angle between v and B is necessarily 90° Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same [2006] (a) 4 (b) 6 (c) 2 (d) 3
F3
S
R F2 A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR, and RQ are F1 , F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is [2008] (a) F3 – F1– F2 (c)
(F3 – F1 )2 – F22
(b)
(F3 – F1 )2 + F22
(d) F3 – F1+F2
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
48.
49.
A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of r induction B . After 3 seconds, the kinetic energy of the particle will be: [2008] (a) 3T (b) 2T (c) T (d) 4T A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction 1 Wb / m 2 in such a way that its axis makes p r an angle of 60° with B .The magnetic flux linked with the disc is: [2008] (a) 0.02 Wb (b) 0.06 Wb (c) 0.08 Wb (d) 0.01 Wb A galvanometer having a coil resistance of 60 W shows full scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents upto 5.0 amp by [2009] (a) putting in series a resistance of 15 W (b) putting in series a resistance of 240 W
(
50.
51.
53.
55.
56.
(c) putting in parallel a resistance of 15 W (d) putting in parallel a resistance of 240 W The magnetic force acting on a charged particle of charge – 2 mC in a magnetic field of 2T acting in y direction, when the particle velocity is 2 ˆi + 3 ˆj ×106 ms –1, is [2009]
(
52.
)
54.
)
(a) 4 N in z direction (b) 8 N in y direction (c) 8 N in z direction (d) 8 N in – z direction Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle: [2009] (a) depends on R and not on v (b) is independent of both v and R (c) depends on both v and R (d) depends on v and not on R A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of f revolutions/s. The value of magnetic induction in Wb/m2 at the centre of the ring is [2010] m0 q f m0 q (a) 2 p R (b) 2p f R m0 q f m0 q (c) (d) 2R 2f R
57.
5 A galvanometer has a coil of resistance 100 ohm and gives a full-scale deflection for 30 mA current. It is to work as a voltmeter of 30 volt range, the resistance required to be added will be [2010] (a) 900 W (b) 1800 W (c) 500 W (d) 1000 W A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the r loop. If the force on one arm of the loop is F , the net force on the remaining three arms of the loop is [2010] r r (a) 3 F (b) – F r r (c) – 3 F (d) F A particle having a mass of 10–2 kg carries a charge of 5 × 10–8C. The particle is given an initial horozontal velocity of 105 ms–1 in the r presence of electric field E and magnetic field r B . To keep the particle moving in a horizontal direction, it is necessary that r (1) B should be perpendicular to the direction r of velocity and E should be along the direction of velocity. r r (2) Both B an d E should be along the direction of velocity. r r (3) Both B and E are mutually perpendicular and perpendicular to the direction of velocity. r (4) B should be along the direction of velocity r and E should be perpendicular to the direction of velocity. Which one of the following pairs of statements is possible? (a) (2) and (4) (b) (1) and (3) (c) (3) and (4) (d) (2) and (3) A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10–4 m2 carries a current of 2.0 A. It suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10–2 tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be: (a) 3 × 10–2 N-m (b) 3 × 10–3 N-m –3 (c) 1.5 × 10 N-m (d) 1.5 × 10–2 N-m
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
6 58.
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-z plane. If the current in the loop is i., the resultant magnetic field due to the two semicircular parts at their common centre is (a)
m0 i
(b)
2R
I1 d
m 0i 2 2R
m 0i m 0i (d) 2R 4R A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, what is the force on the arm AC? [2011]
(c)
59.
63.
A
(a)
60.
61.
B
C
(b)
r -F
r r (c) F (d) 2F A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron [2011] (a) will turn towards right of direction of motion (b) speed will decrease (c) speed will increase (d) will turn towards left direction of motion A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is [2011M]
(a)
S2 (S + G)
(b)
SG (S + G)
G G2 (d) (S + G) (S + G) A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I1 at a distance d from the conductor as shown in figure. The loop will experience [2011M]
(c) 62.
r - 2F
I
64
I (a) a net repulsive force away from the conductor (b) a net torque acting upward perpendicular to the horizontal plane (c) a net torque acting downward normal to the horizontal plane (d) a net attractive force towards the conductor Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is [2011M] m0 q m 0 qf (a) (b) 2f R 2R m0q m 0 qf (c) 2pf R (d) 2pR Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2 I, respectively. The resultant magnetic field induction at the centre will be: [2012] (a)
5m 0 I 2R
(b)
3m0 I 2R
m0 I m0 I (d) 2R R An alternating electric field, of frequency v, is applied across the dees (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by : [2012]
(c)
65.
(a)
B=
mn and K = 2mp2n2R2 e
(b)
B=
2pm n and K = m2pnR2 e
(c)
B=
2pmn and K = 2mp2n2R2 e
mn and K = m2pnR2 B= e Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
(d)
7 66.
67.
68.
An a-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m 2 . The de-Broglie wavelength associated with the particle will be : [2012] (a) 1 Å
(b) 0.1 Å
(c) 10 Å
(d) 0.01 Å
A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an a-particle to describe a circle of same radius in the same field? [2012M] (a) 2 MeV
(b) 1 MeV
(c) 0.5 MeV
(d) 4 MeV
70.
A current loop in a magnetic field [NEET 2013] (a) can be in equilibrium in one orientation (b) can be in equilibrium in two orientations, both the equilibrium states are unstable (c) can be in equilibrium in two orientations, one stable while the other is unstable (d) experiences a torque whether the field is uniform or non-uniform in all orientations
69.
When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a speed v0 it moves with an initial acceleration 3a0 towards west. The electric and magnetic fields in the room are respectively [NEET 2013] (a)
2ma0 ma0 west, ev down e 0
71.
(b)
3ma0 ma0 east, up ev0 e
(c)
3ma0 ma0 east, ev down e 0
(d)
2ma0 ma0 west, ev up e 0
A long straight wire carries a certain current and weber produces a magnetic field of 2 × 10–4 at a m2 perpendicular distance of 5 cm from the wire. An electron situated at 5 cm from the wire moves with a velocity 107 m/s towards the wire along perpendicular to it. The force experienced by the electron will be [NEET Kar. 2013] (charge on electron =1.6 × 10–19 C) (a) Zero (b) 3.2 N (c) 3.2 × 10–16 N (d) 1.6 × 10–16 N A circular coil ABCD carrying a current i is placed in a uniform magnetic field.r If the magnetic force on the segment AB is F , the force on the remaining segment BCDA is [NEET Kar. 2013] A i D
(a) (c)
r F r 3F
B
C (b) (d)
r -F r -3F
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
8
SOLUTIONS 1.
2. 3 4.
(d) The plane of coil will orient itself so that area vector aligns itself along the magnetic field. So, the plane will orient perpendicular to the magnetic field. (b) Tesla is the unit of magnetic field. (b) Energy is stored in magnetic field. df e 1 df ;i= = (c) e = dt R R dt 1 df .dt Total charge induced = ò i dt = ò R dt f
=
1 2 1 d f = (f 2 - f1 ) ò Rf R 1
5.
6.
m I 1 (a) B = 0 Þ B µ 2 pr r As the distance is increased to three times, the magnetic induction reduces to one third. 1 Hence, B = ´ 10 -3 tesla = 3.33 ´ 10 -4 tesla 3 (d) For a charged particle orbiting in a circular path in a magnetic field
mv2 Bqr = Bqv Þ v = r m 2 or, mv = Bqvr Also, EK =
1 2 1 r Bqr B 2 q 2 r 2 mv = Bqvr = Bq . = 2 2 2 m 2m
For deuteron, E1 =
8.
2 2 2
B q r 2m E1 1 50keV 1 = Þ = Þ E2 = 100keV 2 E2 2 E2 mv or r µ v (c) r = qB As v is doubled, the radius also becomes double. Hence, radius = 2 × 2 = 4 cm mi 1 (a) B = 0 or B µ 2pr r When r is doubled, the magnetic field becomes half, i.e., now the magnetic field will be 0.2 T.
For proton, E2 =
7.
B2 q2 r 2 2 ´ 2m
9. (b) F = Bil = 2 ×1.2 × 0.5 = 1.2 N 10. (a) To convert a galvanometer into an ammeter, one needs to connect a low resistance in parallel so that maximum current passes through the shunt wire and ammeter remains protected. 11. (a) A current carrying coil has magnetic dipole r r moment. Hence, a torque pm ´ B acts on it in magnetic field. 12. (a) The force acting on a charged particle in magnetic field is given by r r r F = q ( v ´ B ) or F = qvB sin q, When angle between v and B is 180°, F=0 13. (d) When the deflection produced by electric field is equal to the deflection produced by magnetic field, then the electron can go undeflected. m 0i and so it is independent of 14. (b) B = 2pr thickness. The current is same in both the wires, hence magnetic field induced will be same. 15. (a) Current (I) = 12 A and magnetic field (B) = 3 × 10–5 Wb/m2. Consider magnetic field ur B at distance r.. m I Magnetic field , B = 0 2pr m0 I (4p ´ 10 -7 ) ´ 12 = = 8 ´ 10 -2 m 2 pB 2 ´ p ´ (3 ´ 10 -5 ) 16. (d) According to Biot Savart law, r uur uur m 0 i(d l ´ r ) dB = 4p r3 17. (c) K.E. of electron = 10 eV 1 Þ mv 2 = 10 eV 2 1 Þ (9.1 ´ 10 -31 )v 2 = 10 ´ 1.6 ´ 10 -19 2
Þ r=
2 Þ v =
2 ´ 10 ´ 1.6 ´ 10 -19
9.1 ´ 10 -31 Þ v2 = 3.52 × 1012 Þ v = 1.88 × 106 m
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
9 Also, we know that for circular motion 2
mv mv = 11 cm = Bev Þ r = Be r
B=
2nd Case : l = 2(2pr ¢) Þ r ¢ =
I
18. (a)
C
A O
I D
B
Net magnetic field on AB is zero because magnetic field due to both current carrying wires is equal in magnitude but opposite in direction. 19. (c) The electron moves with constant velocity without deflection. Hence, force due to magnetic field is equal and opposite to force due to electric field. E 20 = = 40 m/s B 0.5 20. (c) Maximum current which can pass through galvanometer, Ig = 0.004A
qvB = qE Þ v =
RW
C
D
I–Ig I A Ig
20W
B
Let R be the resistance of shunt. We know potential drop across AB = Potential drop across CD R (I – Ig) = Ig (20) Þ R (1 – 0.004) = 0.004 × 20 Þ R = 0.08 W 21. (b) In a perpendicular magnetic field, the path of a charged particle is a circle, and the magnetic field does not cause any change in energy. m 0 2i1i2 l - 7 2 ´ 1´ l ´ l ´ = 10 ´ 4p r 1 –7 = 2 × 10 N/m. [This relates to the definition of ampere]
22. (a) F =
8 8 G = 0.8 amp . = =1´ 23. (b) I s = I ´ + 10 2 8 S +G 24. (b) Let l be length of wire.
m0 In m 0 I ´ 2p m0 pI = = [ Q n = 1]…(1) 2r 2l l l 4p
m In 2m I p æ m pI ö B¢ = 0 = 0 = 4 ç 0 ÷ = 4B l l è l ø , 2 4p 2 using (1) (where n = 2) 25. (b) We know that magnetic field at the centre of circular coil, m 0 In 4p ´ 10-7 ´ 2 ´ 50 = = 1.25 ´ 10-4 N 2r 2 ´ 0.5 26. (c) Potential difference (V) = 1V, K.E. acquired = qV = 1.6 × 10–19 × 1 = 1.6 × 10–19 joules = 1 eV 27. (b) Inside a hollow pipe carrying current, the magnetic field is zero, since according to Ampere'slaw, Bi. 2pr = m0 × 0 Þ Bi = 0. But for external points, the current behaves as if it was concentrated at the axis only; so, m i outside, B0 = 0 . Thus, the magnetic field 2 pr is produced outside the pipe only. 28. (d) When the current flows in both wires in the same direction then magnetic field at half way due to the wire P, uur m I m I m Bp = 0 1 = 0 1 = 0 5 p .5 2p 2p 2 (where I1= 2.5 amp) r The direction of B p is downward
B=
P
Q ×
2.5 amp
C
5 amp
×
5m Magnetic field at half way due to wire Q uur m I m BQ = 0 2 = 0 [upward × ] 5 p 2p 2 [where I 2 = 2.5amp. ]
l 2p Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
Ist case : l =2pr Þ r =
10
uur 34. (d) Force due to electric field = qE
Net magnetic field at half way ur ur ur B = B P + BQ = -
m0 m0 m0 + = (upward ) 2p p 2p
Hence, net magnetic field at midpoint = 29. (b) r =
m0 2p
5
mv sin q 3 ´ 10 sin 30 º = Be 0.3 ´ 10 8
3 ´ 105 ´
1 2 = 0.5 ´10 – 2 m = 0.5cm.
3 ´10 7 30. (a) Electron moves undeflected if force exerted due to electric field is equal to force due to magnetic field. ur r |E| r ur ur q | v || B |= q | E | Þ | v |= ur |B|
31. (b) For circular path in magnetic field, mrw2 = qvB 2 Þw =
qvB mr
ur uur Force due to magnetic field = q(v ´ B ) uur ur uur Net force experienced = qE + q(v ´ B ) 35. (b) Magnetic force acts perpendicular to the velocity. Hence speed remains constant. 36. (c) B = m 0 ni
æ nö B1 = (m0 ) ç ÷ (2 i ) = m 0 ni = B è 2ø
Þ B1 = B 37. (c) A high resistance is connected in series so that less current passes through voltmeter. 38. (a) Rg = 50W, Ig = 25 × 4 × 10–AW = 10–2 A Range of V = 25 volts V = Ig(Re + Rg) \ Re =
V - Rg = 2450W Ig
R A
B Ig
As v = rw
Re
Rg
mv v Þrµ qB B r 40. (a) The direction of B is along ( - kˆ ) \ The magnetic force w qB ur r ur n= Þ n= F = Q (v ´ B) = Q (viˆ) ´ B( - kˆ) = QvBjˆ 2p 2 pm r Þ F is along OY. 32. (c) A galvanometer can be converted into a voltmeter by connecting the high 41. (b) Force on a particle moving with velocity v in ur ur ressistance in series with the galvanometer a magnetic field B is F = q (v ´ B) so that only a small amount of current passes r r If angle between v & B is either zero or through it. 180º, then value of F will be zero as cross 33. (a) Let I be current and l be the length of the r r wire. product of v and B will be zero. So option (b) is correct. m In m I ´ p For Ist case : B = 0 = 0 where 42. If R1 & R2 be the radius of the circular wires, 2r l R1 2 2pr = l and n = 1 = . If same potential is applied on them, R2 1 l current in Ist will be half that in the later. If V For IInd Case : l = 2(2pr ') Þ r ' = 4p potential is applied on them, current in them m 0 nI m 0 2 I 4m 0 I p V V B' = = = = 4B = & . l l 2r ' 2R R 2 4p Now magnetic field at the centre of circular Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html q(r w) B qB \ w = Þ w= mr m \ If n is frequency of roatation, then 2
39. (d) r =
11 coil, =
µ0 I 2r
z v x
For first wire, field B1 =
µ0V 2R ´ 2R
For second wire, field B2 =
v
46. (b)
µ0V 2( R / 2) ´ R
Given B1 = B2 The given data do not provide any required result. There is a mistake in the framing of the question. 43. (a) If the electric field is switched off, and the same magnetic field is maintained, the electrons move in a circular orbit and electron will travel a magnetic field ^ to its velocity.. 44. (a) In mass spectrometer, when ions are accelerated through potential V 1 2 mv = qV 2
x l
47.
When a test charge q0 enters a magnetic field r r B directed along z-axis, with a velocity v making angles d with the z-axis. The time period of the motion is independent of R and v. (b) According to the figure the magnitude of force on the segment QM is F3 –F1 and PM is F2.
..........(i)
As the magnetic field curves the path of the ions in a semicircular orbit Bqv =
mv 2 BqR Þv= R m
.......... (ii)
Substituting (ii) in (i) 2
Therefore, the magnitude of the force on
1 é BqR ù m = qV 2 êë m úû
or
q 2V = 2 2 m B R
segment PQ is 48.
(c) When a charged particle enters a transverse magnetic field it traverse a circular path. Its kinetic energy remains constant.
49.
(a) Here, B =
Since V and B are constants,
q 1 \ µ 2 m R 45. (d) Magnetic moment, m = IA =
qv qvR ( pR 2 ) = 2 pR 2
q 2pR ù é êëQ I = T and T = v úû
(F3 – F1 )2 + F22
1 (Wb/m2) p q = 60° Area normal to the plane of the disc
pr 2 2 Flux = B × normal area
= pr 2 cos 60° =
=
0.2 ´ 0.2 = 0.02Wb 2
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
12 50.
(c) G = 60W, Ig=1.0A, I=5A.
60W G
Ig
I
55.
(b) The force on the two arms parallel to the field is zero. <
51.
Ig G I – Ig
=
1 ´ 60 = 15W 5 –1
–F <
S=
56.
Thus by putting 15 W in parallel, the galvanometer can be converted into an ammeter. (d) The magnetic force acting on the charged paraticle is given by r r r F = q v× B
(
)
ˆ ´ 106 } ´ (2ˆj)] = (–2× 10–6) [{2iˆ + 3j)
ˆ = –4(2k)
52.
53.
57.
= –8kˆ \ Force is of 8N along – z-axis. (b) The time period of the charged particle is given by T =
2πm qB
Thus, time period is independent of both v and R. (d) Current in the ring due to rotation, q qw q × 2p f = = 2p T 2p Therefore, magnetic field at the centre of the ring is
I=
58.
\ Force on remaining arms = – F (d) Force due to electric field acts along the direction of the electric field but force due to the magnetic field acts along a direction perpendicular to both the velocity of the charged particle and the magnetic field. Hence both statements (2) and (3) are true. In statement (2), magnetic force is zero, so, electric force will keep the particle continue to move in horizontal direction. In statement (3), both electric and magnetic forces will be opposite to each other. If their magnitudes will be equal then the particle will continue horizontal motion. (d) Torque on the solenoid is given by t = MB sin q where q is the angle between the magnetic field and the axis of solenoid. M = niA \ t = niA B sin 30° 1 = 2000 ´ 2 ´ 1.5 ´ 10 -4 ´ 5 ´ 10 -2 ´ 2 -2 = 1.5 ´ 10 N - m (b) Magnetic fields due to the two parts at their common centre are respectively, m i m i B y = 0 and Bz = 0 4R 4R z i
m0 I m0 q f m 0 q 2p f = B= 2R = R 2R 2 2p
54.
F B
S Let S be the shunt resistance connected in parallel to galvanometer Ig G = (I – Ig) S,
<
<
(I–Ig)
(a) Let the resistance to be added be R, then 30 = Ig (r + R) \
R=
i
30 30 - 100 -r = Ig 30 ´10-3
= 1000 – 100 = 900 W
y i
Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
Resultant field =
B y2 + Bz2 2
æ m iö æ m iö = ç 0 ÷ +ç 0 ÷ è 4R ø è 4R ø
59.
2
m 0i m i = 0 4R 2 2 R (b) Let a current i be flowing in the loop ABC in the direction shown in the figure. If the length of each of the sides AB and BC be x then r | F| = i x B =
13 (c) To keep the main current in the circuit unchanged, the resistance of the galvanometer should be equal to the net resistance. æ GS ö \G = ç + S¢ è G + S ÷ø GS ÞG= S¢ G+S
61.
2.
\ S¢ = I
G2 . G+S I
G
G
A
S
62.
I1
(d) I
F1
F3
Direction of magnetic field where B is the magnitude of the magnetic force. r The direction of F will be in the direction perpendicular to the plane of the paper and going into it. By Pythagorus theorem,
F2 1 , and F3 and F4 are equal d and opposite. Hence, the net attraction force will be towards the conductor. (a) When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is
F1 > F2 as F µ
63.
q = qf T Magnetic field at the centre of the ring is I=
AC = x 2 + x 2 = 2x \ Magnitude of force on AC 2 x B sin 45°
=i 2 xB´
60.
F4
C
B
=i
S¢
1 2
r = ixB = | F | The direction of the force on AC is perpendicular to the plane of the paper and r going out of it. Hence, force on AC = - F r r (b) v and B are in same direction so that magnetic force on electron becomes zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.
B= 64.
m0I m 0 qf = 2R 2R
(a) B2 B1
The magnetic field, due the coil, carrying current I Ampere
m0 I 2R Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html B1 =
14 The magnetic field due to the coil, carrying current 2I Ampere
Ra =
m 0 (2I ) 2R The resultant B B2 =
Bnet = B12 + B22 + 2B1B2 cos q, q = 90°
Bnet = B12 + B22 =
m0 (2I ) 1+ 4 2R
68.
5 m0 I 2R (c) Time period of cyclotron is =
65.
T=
Þ
1 2pm mu p 2pm = = ; B= u; R = u eB eB eB e
P = eBR = e ´
2(4 m) K ' 2qB
R K = Ra K' but R = Ra (given) Thus K = K¢ = 1 MeV (c) A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable. r uur ur Q t = M ´ B = M B sin q If q = 0° Þ t = 0 (stable) If q = p Þ t = 0 (unstable)
2pmu R = 2pmuR e
p 2 (2pmuR )2 = = 2p2mu2R2 2m 2m The de-Broglie wavelength associated with the particle will be
K.E. =
66.
h h Þ l= , p mv h = plank’s constant = 6.63 × 10–34 J.S For circular motion = Fc = qvB
Do not experience a torque in some orientations Hence option (c) is correct.
l=
Þ
m n2 = q nB r
Þ
mv =r qB r=
mv Þ mv = qrB qB
æ 1ö Þ (2e) (0.83 × 10–2) çè ÷ø 4
67.
69.
(a)
When moves with an acceleration a0 towards west, electric field F ma 0 = (West) q e When moves with an acceleration 3a0 towards east, magnetic field
E=
2ma 0 (downward) ev0 -34 6.6 ´ 10 ´4 l= 70. (c) Given: -19 ´ 0.83 ´ 10-2 2 ´ 1.6 ´ 10 Magnetic field B = 2 × 10–4 weber/m2 l = 9.93 × 10–34 + 21 Velocity of electron, v = 107 m/s (b) According to the principal of circular Lorentz force F = qvB sin q motion in a magnetic field = 1.6 × 10–19 × 107 × 2 × 10–4 (Q q = 90°) = 3.2 × 10–16 N mv 2 r r r Fc = Fm Þ = qVB 71. Here, FAB + FBCDA = 0 R r r r Þ FBCDA = - FAB = - F mv P 2 m.k = = Þ R= r qB qB qB (Q FAB = F ) Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html
B=