Q: A conducting solid sphere of radius R and mass M carries a charge Q. The sphere is rotating about an axis passing through its center with a uniform angular speed . The ratio of the magnitudes of the m

aligned & d M=d I A \\ & A= r^2= (R )^2 \\ & d I= d a T = (2 r)(R d ) 2 \\ & d I= 2 R^2 d 2 aligned dI = R ^2 ~d Magnetic dipole moment : M= d M= _0^ R^2 R^2 ^3 d M= R^4 _0^ ^3 d ( _0^ ^3 d = 4 3 ) M = ( Q 4 R ^2 ) R ^4 ( 4 3 ) Magnetic dipole moment M = QR ^2 3 Angular momentum L = ( 2 5 MR ^2 ) M ~L = QR ^2 3 2 5 MR ^2 = Q 2 M ( 5 3 ) = 5 3 =1.67

Q: A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass m and radius r and it is in a uniform ve

aligned & Let loop makes angle with vertical. \\ & in equilibrium _ net =0 \\ & _0= MB (90- )- mg . r =0 \\ & I. r ^2 B _0 = mg r.sin \\ & = rIB _0 mg \\ & aligned

Q: A positive, singly ionized atom of mass number A_ M is accelerated from rest by the voltage 192 ~V . Thereafter, it enters a rectangular region of width w with magnetic field B _0=0.1 k Tesla, as show

aligned & x =2 R \\ & x =2 P qB x = 2 2 mqV qB x = 2 ~B 2 mV q aligned Option A For H ^ + m = 5 3 10^ -27 ~kg x= 2 0.1 2 5 3 10^ -27 192 1.6 10^ -19 =4 ~cm Option B For A _ m =144 x= 2 0.1 2 144 5 3 10^ -27 192 1.6 10^ -19 =48 ~cm Option C for A _m=1 x =4 ~cm \& for A _ m =196 x =56 ~cm . so x _0=4 ~cm \& x _1=56 ~cm x _1- x _0=52 ~cm . Option D Minimum width =R for A _ M =196 aligned & R = P qB = 2 mqV qB \\ & R = 1 ~B 2 mV q \\ & w _ = R = 1 0.1 2 196 5 3 10^ -27 192 1.6 10^ -19 =28 ~cm aligne

Q: An infinitely long wire, located on the z-axis, carries a current I along the +z-direction and produces the magnetic field B . The magnitude of the line integral B d l along a straight line from the p

aligned & | d |= rd \\ & | B |= _0 I 2 r \\ & B d = | B || d | 0^ \\ & = ( _0 I 2 r ) ( rd ) \\ & = _ _1 ^ _2 _0 I 2 d = _0 I 2 [ _2- (- _1 ) ] aligned [ _1 is anticlockwise hence taken negative] aligned & _1= a 3 a = 3 \\ & _1= 3 \\ & _2= a a =1 \\ & _2= 4 \\ & B d = _0 I 2 [ 3 + 4 ] \\ & = 7 _0 I 24 \\ & Ans. Option (1) aligned

Question 1

A hollow, right circular cone of base radius R and height h, with its tip at the origin is rotating about the Z-axis with an angular velocity , as shown in the figure. The cone carries a total charge

Accepted Answer

For a point on the axis at a distance z R and z h, the rotating cone can be considered as a magnetic dipole. The magnetic field on the axis of a dipole is given by: B = _0 4 2M z^3 where M is the magnetic dipole moment of the cone. To find M, consider an elemental ring on the cone at a vertical distance y from the origin, with vertical thickness dy. The radius of this ring is r = R h y and its slant length is dl = R^2+h^2 h dy. The surface charge density of the cone is = Q R R^2+h^2 . The charge

Question 2

In a vacuum chamber, a particle of charge 1\ C and mass 1 mg is projected with a velocity ( i + 2 j ) ms^ -1 from the XZ plane at time t = 0 in an electric field of 1 i Vm^ -1 . At t = 0.2 s, the elec

Accepted Answer

For the time interval 0 t 0.2 s, the particle is under the influence of the electric field and gravity. The charge-to-mass ratio of the particle is q m = 10^ -6 10^ -6 = 1 C kg^ -1 . The acceleration of the particle is: a = q E m + g = 1( i ) - 10 j = i - 10 j ms^ -2 Given the initial velocity v _0 = i + 2 j ms^ -1 and initial vertical position y_0 = 0 (since it is projected from the XZ plane), we can find the velocity and position at t = 0.2 s. Velocity at t = 0.2 s: v (0.2) = v _0 + a t = ( i

Question 3

A conducting solid sphere of radius R and mass M carries a charge Q. The sphere is rotating about an axis passing through its center with a uniform angular speed . The ratio of the magnitudes of the m

Accepted Answer

aligned & d M=d I A \ & A= r^2= (R )^2 \ & d I= d a T = (2 r)(R d ) 2 \ & d I= 2 R^2 d 2 aligned dI = R ^2 ~d Magnetic dipole moment : M= d M= _0^ R^2 R^2 ^3 d M= R^4 _0^ ^3 d ( _0^ ^3 d = 4 3 ) M = ( Q 4 R ^2 ) R ^4 ( 4 3 ) Magnetic dipole moment M = QR ^2 3 Angular momentum L = ( 2 5 MR ^2 ) M ~L = QR ^2 3 2 5 MR ^2 = Q 2 M ( 5 3 ) = 5 3 =1.67

Question 4

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass m and radius r and it is in a uniform ve

Accepted Answer

aligned & Let loop makes angle with vertical. \ & in equilibrium _ net =0 \ & _0= MB (90- )- mg . r =0 \ & I. r ^2 B _0 = mg r.sin \ & = rIB _0 mg \ & aligned

Question 5

A positive, singly ionized atom of mass number A_ M is accelerated from rest by the voltage 192 ~V . Thereafter, it enters a rectangular region of width w with magnetic field B _0=0.1 k Tesla, as show

Accepted Answer

aligned & x =2 R \ & x =2 P qB x = 2 2 mqV qB x = 2 ~B 2 mV q aligned Option A For H ^ + m = 5 3 10^ -27 ~kg x= 2 0.1 2 5 3 10^ -27 192 1.6 10^ -19 =4 ~cm Option B For A _ m =144 x= 2 0.1 2 144 5 3 10^ -27 192 1.6 10^ -19 =48 ~cm Option C for A _m=1 x =4 ~cm \& for A _ m =196 x =56 ~cm . so x _0=4 ~cm \& x _1=56 ~cm x _1- x _0=52 ~cm . Option D Minimum width =R for A _ M =196 aligned & R = P qB = 2 mqV qB \ & R = 1 ~B 2 mV q \ & w _ = R = 1 0.1 2 196 5 3 10^ -27 192 1.6 10^ -19 =28 ~cm aligne

Question 6

An infinitely long wire, located on the z-axis, carries a current I along the +z-direction and produces the magnetic field B . The magnitude of the line integral B d l along a straight line from the p

Accepted Answer

aligned & | d |= rd \ & | B |= _0 I 2 r \ & B d = | B || d | 0^ \ & = ( _0 I 2 r ) ( rd ) \ & = _ _1 ^ _2 _0 I 2 d = _0 I 2 [ _2- (- _1 ) ] aligned [ _1 is anticlockwise hence taken negative] aligned & _1= a 3 a = 3 \ & _1= 3 \ & _2= a a =1 \ & _2= 4 \ & B d = _0 I 2 [ 3 + 4 ] \ & = 7 _0 I 24 \ & Ans. Option (1) aligned

Question 7

Which one of the following options represents the magnetic field B &#8594;  at O due to the current flowing in the given wire segments lying on the x y  plane?

Accepted Answer

Magnetic field due to a straight wire will be zero if the line of the wire passes through the point of interest. Therefore, In the above figure, B B C + B D E + B F G = 0 Therefore, B &#8594; net &#160; = B &#8594; A B + B &#8594; C D &#160; + B &#8594; E F Magnetic field due to a finite wire is given by, B = &#956; 0 I 4 &#960; d sin &#945; + sin &#946; Using above formula, B &#8594; A B = &#956; 0 I 4 &#960; L &#215; 1 2 - k ^ Magnetic field due to a circular arc is, B = &#956; 0 I 2 R &#215;

Question 8

A small circular loop of area A  and resistance R  is fixed on a horizontal x y -plane with the center of the loop always on the axis n ^ of a long solenoid. The solenoid has m  turns p

Accepted Answer

Note: This question was given bonus by JEE council. The magnetic field in a solenoid is given by,  B &#8594; = &#956; 0 m I n ^ . Therefore, the flux through the loop will be, &#981; = A k ^ &#183; &#956; 0 m I n ^ For case I: &#981; = B A 2 cos &#969; t The emf will be, &#949; = B A &#969; 2 sin &#969; t Therefore, the current in the loop will be, i = B A &#969; 2 R sin &#969; t The magnetic moment of the loop will be, m &#8594; = i A k ^ = B A 2 &#969; 2 R sin &#969; t k ^ Therefore, the

Question 9

Two concentric circular loops, one of radius  R  and the other of radius  2 R , lie in the  x y -plane with the origin as their common center, as shown in the figure. The smaller l

Accepted Answer

(A) Magnetic field at the plane of the ring is perpendicular to the plane. However, they bend as they move forward. (B) By symmetry, we can say that B &#8594; will be same at all the points having the same radial distance. so B &#8594; x , y will depend only the radial distance r = x 2 + y 2 (C) B net &#160; centre &#160; = &#956; 0 I 2 2 2 R - &#956; 0 I 1 2 R = &#956; 0 4 R I 2 - 2 I 1 Since, I 2 &#62; 2 I 1 so B net at the centre will be non-zero in &#8855; direction, But at some other point,

Question 10

Paragraph A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a co

Accepted Answer

For a super conducting loop, net flux passing through it will remain constant. &#981; total &#160; i = &#981; total &#160; f As area is same, therefore, &#956; 0 m 2 &#960;r 3 = &#956; 0 I 2 a &#8658; I = a m &#960;r 3 &#8658; I &#8733; m r 3 The current carrying superconducting loop will also behave like a magnet, whose magnetic dipole moment is given by, m 1 = N I A = ( 1 ) a m &#960;r 3 &#215; &#960; a 2 = m a 3 r 3 &#8658; &#160;&#160; m 1 &#8733; m r 3 The repulsive force felt on

Question 11

An  &#945; -particle (mass  4  amu) and a singly charged Sulphur ion (mass  32  amu) are initially at rest. They are accelerated through a potential  V  and then all

Accepted Answer

When a particle moves on a circular path in a magnetic field then, &#8757; R = m v q B = 2 m E q B = 2 m q V q B = 2 m V q B According to given data, &#8757; V = constant &#38; &#160; B = constant &#8756; R &#8733; m q R &#945; R S = m &#945; q &#945; &#215; q S m S = 4 2 &#215; 1 32 &#8658; R &#945; R S = 1 16 = 1 4 Then,  R S R &#945; = 4

Question 12

A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being verti

Accepted Answer

&#964; = M &#183; B &#183; sin &#160; 90 &#176; &#964; = N I &#960; R 2 B 0 &#916; L = &#8747; 0 t &#964; &#183; d t = N &#183; &#960; R 2 &#183; B &#8747; 0 t I &#160; d t ; &#916; L = N &#183; &#960; R 2 &#183; B Q .

Magnetic Effects of Current — Previous Year Questions & Solutions