[PDF] Electricity and Magnetism IIT-JAM 2005-2019

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[PDF] Electricity and Magnetism IIT-JAM 2005-2019 - fiziks

loop When it is pulled out of the magnetic field, a total charge of 5 2 10 Q C - = × flows through the coil The magnitude of magnetic field B is (a) 3 1 10 T

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theory of electromagnetism, and having shewn how this theory may be 1 3-5 The three fundamental units Length, Time and Mass 2, 3 6 Derived units 5 7 electrified It is the established practice of men of science tocall the vitreous

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[PDF] Electricity and Magnetism IIT-JAM 2005-2019 - fiziks

105_84_ElectricityandMagnetism.pdf fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 1

Electricity and Magnetism

IIT-JAM 2005

Q1. A small loop of wire of area

0.01Am, 40Nturns and resistance 20R is

initially kept in a uniform magnetic field

B in such a way that the field is normal to the

loop. When it is pulled out of the magnetic field, a total charge of 5 210QC
flows through the coil. The magnitude of magnetic field B is (a) 3 110T
(b) 3 410T

Ans.: (a)

Solution: Magnetic flux through the loop

NBA

Induced e.m.f

dtd and induced current dtdQ dtd Ri I1 dQdR 1.

Thus ,

10201.040201

BTB 3 101
.

Q2. Two point charges

1 qand 2 q are fixed with a finite distance d between them. It is desired to put a third charge 3 q in between these two charges on the line joining them so that the charge 3 q is in equilibrium. This is (a) possible only if 3 qis positive (b) possible only if 3 qis negative (c) possible irrespective of the sign of 3 q (d) not possible at all

Ans. : (c)

Solution: If

3 q is positive, If 3 q is negative, In both case there is possibility that charge 3 q may be in equilibrium. 1 q 3 q 2 F 1 F d 3 q 2 q 1 q 3 q 1 F 2 F d 3 q 2 q fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 2

IIT-JAM 2006

Q3. Two electric dipoles

P and

2 P are placed at 0,0,0 and 1, 0, 0 respectively with both of them pointing in the zdirection. Without changing the orientations of the dipoles 2 P is moved to

0,2,0. The ratio of the electrostatic potential energy of the dipoles after

moving to that before moving is (a) 1

16 (b)

21 (c) 41 (d) 81

Ans. : (d)

Solution: Electrostatic potential energy

3 1 r U 81
3 23
1 12 rr UU Q4. A small magnetic dipole is kept at the origin in the -xy plane. One wire 1

L is located at

az in the -xz plane with a current I flowing in the positive x direction. Another wire 2 L is at az in -yz plane with the same current I as in 1

L, flowing in the

positive y-direction. The angle made by the magnetic dipole with respect to the positive x-axis is (a) 0

225 (b)

120 (c)

45 (d)

0 270

Ans.: (a)

Solution: Magnetic field at

0z due to wire at azisyBBˆ.

Magnetic field at

0z due to wire at azisxBBˆ.

Resultant magnetic field at

0z makes an angle of

45 withxˆ and

225 with xˆ.

fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 3

IIT-JAM 2007

Q5. A uniform and constant magnetic field B coming out of the plane of the paper exists in a rectangular region as shown in the figure. A conducting rod

PQ is rotated about O with a

uniform angular speed in the plane of the paper. The emf PQ E induced between P and Q is best represented by the graph (a) (b) (c) (d)

Ans.: (a)

Solution: When point

P is inside due to motional emf, potential PQ is positive. When point Q is inside potential

QP is positive or potential PQ is negative.

IIT-JAM 2008

Q6. If the electrostatic potential at a point,xy is given by 24Vxyvolts, the electrostatic energy density at that point 3 /in J m is (a) 0

5 (b)

10 (c)

20 (d)

2 0

4221yx

Ans.: (b)

B P O Q tO PQ E t O PQ E t O PQ E t O PQ E fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 4

Solution: yxVEˆ4ˆ220 /EVm

Electrostatic energy density

3 002 0 /102021

21mJE

IIT-JAM 2009

Q7. An oscillating voltage

0 cosVt V t is applied across a parallel plate capacitor having a plate separation d. The displacement current density through the capacitor is (a) dtV cos 00 (b) dtV cos 000 (c) dtV cos 000 (d) dtV sin 00

Ans.: (d)

Solution: Displacement current density

dtV ttV dtEJ d sin 000 0

Q8. An electric field

IITED

ˆcossinˆrrE

exists in space. What will be the total charge enclosed in a sphere of unit radius centered at the origin? (a) 0

4 (b)

4 (c)

4 (d)

0 4

Ans.: (a)

Solution:

02 00

4ˆsinˆcossinˆ

rddrradEQ enc tVtVcos 0 d fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 5

IIT-JAM 2010

Q9. The magnetic field associated with the electric field vector jtkzEEˆsin 0 is given by (a) itkzcEBˆsin 0 (b) itkzcEBˆsin 0 (c) jtkzcEBˆsin 0 (d) ktkzcEBˆsin 0

Ans.: (a)

Solution:

ˆˆsinkz E kz t jkEB

ˆˆsin sinkE Ekz t i kz t ic

Q10. Assume that

0z plane is the interface between two linear and homogeneous dielectrics

(see figure). The relative permittivities are 5 r for

0z and 4

r for0z. The electric field in the region

0z ismVkkjiE

ˆ4ˆ5ˆ3

1. If there are

no free charges on the interface, the electric field in the region

0z is given by

(a) mVkkjiEˆˆ45ˆ432 (b)mVkkjiE

ˆˆ5ˆ3

2 (c) mVkkjiE

ˆ5ˆ5ˆ3

2 (d) mVkkjiE

ˆ5ˆ5ˆ3

Ans.: (d)

Solution:

12 2

ˆˆ35

EE E i j

and 0 f kkEEDDˆ5ˆ445 1 21
221
mVkkjiEˆ5ˆ5ˆ32 z 0z 4 r 5 r fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 6 Q11. A closed Gaussian surface consisting of a hemisphere and a circular disc of radiusR, is placed in a uniform electric field E , as shown in the figure. The circular disc makes an angle 0

30 with the vertical. The flux of the electric field vector coming out of the

curved surface of the hemisphere is (a) 2 1 2

RE

(b) 2 3 2 RE (c) 2 RE (d) 2 2RE

Ans.: (b)

Solution:

xEzExEzEEˆ21ˆ23ˆ30sinˆ30cos rddRxEzEadE SE

ˆsinˆ21ˆ23

2 G 2/ 02 0 2 sincossin21cos23 ddEER E 2/ 02 0 2 sincos23 ddER E 2/ 02 0 22
cossin21 ddER ERER E 22

23021223

OR 02 cos30 E S

Eda E R

2 3 2 RE E zˆ xˆ 0 30
E fiziks InstituteforNET/JRF,GATE,IITͲJAM,M.Sc.Entrance,JEST,TIFRandGREinPhysics H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016

Phone: 011-26865455/+91-9871145498

Website: www.physicsbyfiziks.com | Email: fiziks.physics@gmail.com 7

IIT-JAM 2011

Q12. Equipotential surface corresponding to a particular charge distribution are given by 222
42
i xyzV , where the values of i

V are constants. The electric field E

at the origin is (a) 0E (b) ˆ2Ex (c) ˆ4Ey (d) ˆ4Ey

Ans.: (d)

Solution:

ˆˆ ˆˆ8 2 2 2 0,0,0 4EVxxyyzzE y

IIT-JAM 2012

Q13. A parallel plate air-gap capacitor is made up of two plates of area 2

10cm each kept at a

distance of

0.88mm. A sine wave of amplitude 10V and

frequency

50Hz is applied across the capacitor as shown in the

figure. The amplitude of the displacement current density (in 2 /mA m) between the plates will be closest to (a)

0.03 (b) 0.30 (c) 3.00 (d) 30.00

Ans.: (a)

Solution: Displacement current density,

dtV ttV dtEJ d sin 000 0 Amplitude of the displacement current density (in mA/m 2 ) , 00 00 0 2 d

VfVJdd

20 0095

1501040.03/291028810

d fVJmAmd u

[PDF] Electricity and Magnetism IIT-JAM 2005-2019 - fiziks

Phone: 011-26865455/+91-9871145498

Electricity and Magnetism

IIT-JAM 2005

Q1. A small loop of wire of area

0.01Am, 40Nturns and resistance 20R is

B in such a way that the field is normal to the

Ans.: (a)

Solution: Magnetic flux through the loop

Induced e.m.f

Thus ,

10201.040201

Q2. Two point charges

Ans. : (c)

Solution: If

Phone: 011-26865455/+91-9871145498

IIT-JAM 2006

Q3. Two electric dipoles

P and

0,2,0. The ratio of the electrostatic potential energy of the dipoles after

16 (b)

21 (c) 41 (d) 81

Ans. : (d)

Solution: Electrostatic potential energy

L is located at

L, flowing in the

225 (b)

120 (c)

45 (d)

Ans.: (a)

Solution: Magnetic field at

0z due to wire at azisyBBˆ.

Magnetic field at

0z due to wire at azisxBBˆ.

Resultant magnetic field at

0z makes an angle of

45 withxˆ and

225 with xˆ.

Phone: 011-26865455/+91-9871145498

IIT-JAM 2007

PQ is rotated about O with a

Ans.: (a)

Solution: When point

QP is positive or potential PQ is negative.

IIT-JAM 2008

5 (b)

10 (c)

20 (d)

4221yx

Ans.: (b)

Phone: 011-26865455/+91-9871145498

Solution: yxVEˆ4ˆ220 /EVm

Electrostatic energy density

21mJE

IIT-JAM 2009

Q7. An oscillating voltage

Ans.: (d)

Solution: Displacement current density

Q8. An electric field

ˆcossinˆrrE

4 (b)

4 (c)

4 (d)

Ans.: (a)

Solution:

4ˆsinˆcossinˆ

Phone: 011-26865455/+91-9871145498

IIT-JAM 2010

Ans.: (a)

Solution:

ˆˆsinkz E kz t jkEB

ˆˆsin sinkE Ekz t i kz t ic

Q10. Assume that

0z plane is the interface between two linear and homogeneous dielectrics

0z and 4

0z ismVkkjiE

ˆ4ˆ5ˆ3

1. If there are

0z is given by

ˆˆ5ˆ3

4 (c)

4 (d)