Download JEE ADVANCE - 7 ANAND(Solutions)

JEEADVANCE- 7ANAND (Solutions)
PART - I: PHYSICS SOLUTION
Single Correct Answer Type
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
1.
Current flows through a straight cylindrical conductor of radius r. The current distributed
uniformly over it cross-section. The magnetic field at a distance x from the axis of the
conductor has magnitude B.
(c) B = r
(d) B = x
(a) B = 0 at the axis
(b) B  
sol.
(a)

B . dl  0
ix 2
r2
B2x  0
2.
ix 2
r2
A uniform magnetic field is directed out of the page. A charged particle, moving in the
plane of the page. Follows as clockwise spiral of decreasing radius as shown. A reasonable
explanation is
(a)
(b)
(c)
(d)
sol.
The
The
The
The
change
change
change
change
is
is
is
is
positive and slowing down
negative and slowing down
negative and speeding up
negative and speeding up
(a) r  qB
mv
r = decreases nd hence
v = decreases
3.
magnetic foce F  qv  B
must be in ward the spiral
q is positive

Time constant for the circuit is
(a) RC
sol.
(b)
2RC
(c)
RC
2
(d)
RC
4
(c)
Page # 1
4.
Sol
A conducting circular loop of radius r carries a constant I. It is placed in a uniform mag

netic field B and that B is perpendicular to the plane of the loop. The magnetic force on
the loop is
(c) Zero
(d) 2IrB
(a) BIr
(b) 2 IrB
(c)
SECTION-B
Multiple Correct Answer(s) Type
1.
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE or MORE are correct.
A particle of charge + q and mass m moving under the influence of a uniform electric field
ˆ
Eiˆ and uniform magnetic field Bk follows a trajectory from P to Q as sown in figure. The
ˆ .Which of the following statements is/are correct?
velocities at P and Q are viˆ and –2vj
(a) E 
3  mv2 


4  qa 
(b) Rate of word done by the electric field at P is
3  mv3 


4 a 
(c) Rate of work done by electric field at P is zero
(d) Rate of work done by both fields at Q is zero
Sol.
(a, b, d)
Work doen by magnetic force is always zero. Therefore, from work energy thoerm:
Work done by electrostatic force from P to Q = kinetic energy at Q kinetic energy at P

 qE  2a  

3  mv2 
E 

4  qa 


1
3
m 4v 2  v2  mv 2
2
2
Rate of work done by mgnetic field (power ) is always zero.
Rate of work done by electric field at P
 3 mv2 
3 mv3
Fv  qEv  q 
   0
v 
4 a
 4 qa 
and rate of word done by electric field at Q = 0 as    90º 
Here,   a ngle between Fe and v

2.

A long straight conductor carrying a current i, is bent to form an almost complete cicular
loop of radius R on it, the magnetic field at the centre of the loop
Page # 2
(c) has magnitude 0i /2R 1  1/  
(a) is directed into the paper
Sol.
(b)
(d)
has magnitude 0i /2r 1  1/  
is directed out of the paper
(a, c)
0l
I
 0 
2R
2R
A flat circular coil, carrying a current, has a magnetic moment 
(a)  has only magnitude; it does not have direction
(b) The direction of  is along the normal to the plane of the coil
(c) The direction of  depends on the direction of the current flow
(d) The direction of  deos not change if the current in the coil is reversed
B
3.
Sol.
4.
Sol.
(b, c)


  N lA
n unit vector perpendicular to A
A charged particle moves in a gravity-free space without change in velocity. Which of the
following is/are possible?
(a) E = 0, B = 0
Sol. (a, b, d)
(b)
E = 0, B  0
(c)
E  0, B  0
(d) E  0, B  0
F  0 if v 
5.
Sol.
E
B
If a charged particle goes unaccelerated in a region cootaining electric and magnetic
fields,


(a) E must be perpendicular to B


(b) v must be perpendicular to E


(c) v must be perpendicular to B
(d) E must be equal to vB
(a, b, c, d)
F  0 if v 
E
B
SECTION-C
(Passage single-correct answer type)
This section contains 2 paragraph. each describing theory, experiment, date etc. Six questions relate to two
paragraphs with Three questions on each paragraph. Each question of a paragraph has only one correct answer
among the four choices (A), (B),(C),(D).
Passage-1 (Question No. 1 & 3)


r
r
If a charged particle is projected with a velocity v in a magnetic field B then force acting on the charged particle

 
is given by Fm  q v  B . This force will change the velocity of particle. now suppose that a current carrying wire



 

is placed in a magnetic field then a small current carrying element experiences a force dF  I dl  B . Where dl
1.
is a vector along the length of element and its direction is along the direction of current flow. Now answer the
following questions.
A charge particle is projected with some velocity in a uniform magnetic field only (no electric field in region)
Page # 3
Sol.
2.
Sol.
3.
Sol.
(a) Particle must follow a circular path
(b) Particle must follow a straight line path
(c) Paticle may follow a straight line path
(d) Velocity of particle must change during its motion
(c)
Suppose particle is projected from point A. After some time particle is at point B. Kinetic energy of particle is same
at points A and B because
(a) Magnetic field is perpendicular to line AB
(b) Magnetic field is always perpendicular to line AB
(c) Intanteneous power produced by magnetic force on particle is zero
(d) Kinetic energy of particle at point A and B cant be same
(c)
A straight current carrying wire is placed in a uniform magnetic filed. The magnetic force on a wire will displace the
wire so
(a) Kinetic energy of wire changes if wire is free to move
(b) Kinetic energy of wire cant change because mangetic force cant do any work
(c) Magnetic force will not act because wire is straight
(d) None of the above
(a)
Passage - 2 (Question No. 4 & 6)
 B.dl
 
Amperes law gives a method to calculate the magnetic field due to given current distribution.The circulation
of the resultant magnetic field along a closed, plane curve is equal to 0 times the total current crossing the area
bounded by the closed curve provided the electric field inside the loop remians constant.
Thus,
4.
 B.dl   i
0
 i2 
The contribution of current i3 to the magnetic field cancles out because the integration is made arounhd the full loop.
Consider a co-axial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer
radii b and c and c respectively. The inner wire carries a current i and outer shell carries an equal and opposite
current. The magnetic field at a distance x from axis for x < a is
(a) Zero
Sol.
1
(b)
0i
2 x
(c)
0ix2
2 a 3
(d)
 0ix
2 a 2
(d)
Consider circle of radius x such that 0 < x < a
 
B . dl  B  2x   0i0

Where i0 is the current encloed within the circle
i0 
i
ix 2
2


x

a 2
a2
Page # 4
0ix 2
 ix
 2x  B  2  B  0 2
a
2a

5.
Sol.
The magnetic field at a distance x from the axis where a < x < b is
0i
(a)
2 x
(c)

2 b 2  a2
0i
2x
The magnetic field at a distance x from the axis where b < x < c is

(d)
zero
B  2x    0i  B 

2 x c

a 

2 x c
0i c2  b 2
(a)
Sol.
(b)
0ix
(a)
Consider circle of radius x such that a < x < b. The current enclosed within the circle is i.

6.
0i
2 b  a 
2

a 
0i c 2  x 2
(b)
2
2
2

2 x c

b 
0i c 2  x 2
(c)
2
2
(d)
zero
(c) Consider a circle of radius x such that b < x < c. The net current throughh the circular
region

i
c
  i c  x 
 c  b 
i x 2  b2
2
 b2
2
2

2
2
  B   i c  x 
2x  c  b 
from Amperes law:
 2x  B 
 0i c 2  x 2
c 2  b
2
2
0
2
2
SECTION-D
(Matrix-Match Type)
This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and four
statements (p, q, r, s ) in Column II. Any given statement in Column I can have correct matching with ONE or MORE
statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given q and
r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.
1.
Match the following
Column I
Column II
(A)
Electric field
(p)
Stationary charge
(B)
Magnetif field
(q)
Moving charge
(C)
Electric force
(r)
Changes the kinetic energy
(D)
Magnetic force
(s)
Does not change kinetic energy
Sol.
(a)  (p, q, r, s)
(b)  (q, s)
(c)  (p, q, r, s)
(d)  (q, s)
2.
A square wire frame ABCD is made of four thin uniform rods of length a each. The charge per unit length on each
of four rods is uniform and  . The frame moves in x-y plane (z = 0 plane) with constant velcoity v  v 0iˆ ,the centre
of frame always lies on x-axis and side BC is parallel to y-axis. At the instant centre of frame is at origin, match the
positions in column-I with respective nature of fields in column II.

Page # 5
Sol.
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
Column I
At point P whose coordinates are (4a, 0, 0)
At point Q whose coordinates are (0, 4a, 0)
At point R whose coordinates are (0, 0, 4a)
At origin
(p,
s)

 (q, s)
 (q, s)
 (p, r)
(p)
(q)
(r)
(s)
Column II
Magnitude of magnetif field is zero
Magnitude of magnetic field is nonzero
Magnitude of electrostatic field is zero
Magnitude of electrostatic field is nonzero
Section -E
(I nteger answer correct Type)
T his sect ion cont ains 5 questions. T he answer t o each of t he quest ions is a single-digit integer,
ranging from 0 t o 9. T he bubble corresponding to t he correct answer is t o be darkened in the ORS.
1.
Two parallel wires P and Q placed at a separation d = 6 cm carry electric currents i1 = 5A and i2 = 2A in oppo
site directions as shown in figure. Find the point on theline PQ where the resultant mangentic field is zero. (in cm
form Q)
Sol.
4
2.
The magnetic field B due to a current carrying circular loop of radius 12 cm at its centre is 0.50 × 10–4 T. The
mangetic field due to this loop at a point on the axis at a distance of 5.0 cm from the centre is x × 10–5 T. Find the
value of x. (round of as integer)
4
Sol.
Page # 6
3.
Sol.
4.
Sol.
5.
Sol.
A long copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic

field B due to this current. (in mT)
5
A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10–4 T parallel
to the current. The magnitude of the resultant mangetic field at a point 2.0 cm away from the wire is x × 10–4 T. Find
the value of x.
5
A long, vertical wire carrying a current of 10 A in the upward direction is placed in a region where a horizontal
magnetic field of magnitude 2.0 × 10–3 T exists from south to north. The point where the resultant magnetic field is
zero is x mm west to the wire. Find the vlaue of x.
1
Page # 7