JEE Advanced PYQ Physics – Moving Charges and Magnetism Mixed Concept Problems

Q 1 :

A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is [2010]

$I B L$
$\frac{I B L}{π}$
$\frac{I B L}{2 π}$
$\frac{I B L}{4 π}$

1

2

3

4

(3)

Let us consider an elemental length $d l$ subtending an angle $d θ$ at the centre of the circle $' O'$ . Let $F_{B}$ be the magnetic force acting on this length.

$∴$ $F_{B} = B I (d l) (upwards)$

$= B I (R d θ)$ $[∵ angle d θ = \frac{arc (d l)}{radius R}]$

$= B I (\frac{L}{2 π}) d θ$ $[∵ 2 π R = L \Rightarrow R = \frac{L}{2 π}]$

Let $T$ be the tension in the wire acting along both ends of the elemental length.

At equilibrium, $2 T \sin (\frac{d θ}{2}) = B I \frac{L}{2 π} d θ$

$\Rightarrow 2 T (\frac{d θ}{2}) = B I \frac{L}{2 π} d θ$ $[∵ \frac{d θ}{2} is very, very small]$

$∴$ $T = \frac{B I L}{2 π}$

Q 2 :

An $α$ -particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $α$ -particle and the sulfur ion move in circular orbits of radii $r_{α}$ and $r_{s}$ , respectively. The ratio $(\frac{r_{s}}{r_{α}})$ is __________. [2021]

(4)

$Here, force F = q V B is balanced by centripetal force$

$F_{e} = \frac{m V^{2}}{r}$

$∴$ $q V B = \frac{m V^{2}}{r}$

or $r = \frac{m V}{q B}$ $= \frac{\sqrt{2 m q V}}{q B}$

$\frac{p^{2}}{2 m} = K.E. = q V$

$p = \sqrt{2 m q V}$

Therefore, $r = \frac{p}{q B} = \frac{\sqrt{2 m q V}}{q B}$

$\frac{r_{s}}{r_{α}} = \sqrt{\frac{m_{s}}{q_{s}}} \times \sqrt{\frac{q_{α}}{m_{α}}}$ $= \sqrt{\frac{32}{1}} \times \sqrt{\frac{2}{4}} = 4$

$∴$ $\frac{r_{s}}{r_{α}} = 4$

Q 3 :

In the $x y$ -plane, the region $y > 0$ has a uniform magnetic field $B_{1} \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_{2} \hat{k} .$ A positively charged particle is projected from the origin along the positive $y$ -axis with speed $v_{0} = π {ms}^{- 1}$ at $t = 0$ , as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$ -axis from below for the first time. If $B_{2} = 4 B_{1},$ the average speed of the particle, in ${ms}^{- 1}$ , along the $x$ -axis in the time interval $T$ is _______. [2018]

(2)

$Average speed along x -axis,$ $v_{x} = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

Here, $r_{1} = \frac{m v_{0}}{q B_{1}}$ and $r_{2} = \frac{m v_{0}}{q B_{2}}$

$∵$ $B_{1} = \frac{B_{2}}{4}$ $∴$ $r_{1} = 4 r_{2}$

Time spent by charged particle in $B_{1}$ , $t_{1} = \frac{π m}{q B_{1}}$

Time spent by charged particle in $B_{2}$ , $t_{2} = \frac{π m}{q B_{2}}$

Total distance along $x$ -axis,

$d_{1} + d_{2} = 2 r_{1} + 2 r_{2} = 2 (r_{1} + r_{2}) = 2 (5 r_{2}) = 10 r_{2}$

$Average speed = \frac{10 r_{2}}{5 t_{2}} = 2 \frac{m v_{0}}{q B_{2}} \times \frac{q B_{2}}{π m}$ $= \frac{2 v_{0}}{π}$

$= 2 {ms}^{- 1}$ $(∵ v_{0} = π {ms}^{- 1})$

Q 4 :

A long circular tube of length 10 m and radius 0.3 m carries a current $I$ along its curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I = I_{0} \cos (300 t)$ where $I_{0}$ is constant. If the magnetic moment of the loop is $N μ_{0} I_{0} \sin (300 t),$ then $' N'$ is ________. [2011]

(6)

Consider the circular tube as a long solenoid. The wires are closely wound. Magnetic field B inside the solenoid $B = μ_{0} n i$

$∴$ $B = \frac{μ_{0} I}{L}$ $[∵ n i = \frac{I}{L}]$

Flux passing through the circular coil, $ϕ = B A$ $= (\frac{μ_{0} I}{L}) (π r^{2})$

Induced emf, $e = - \frac{d ϕ}{d t} = - (\frac{μ_{0} π r^{2}}{L}) \frac{d I}{d t}$

Induced current, $i = \frac{e}{R} = - (\frac{μ_{0} π r^{2}}{L R}) \frac{d I}{d t}$

Magnetic moment, $M = i A = i π r^{2}$

or $M = - (\frac{μ_{0} π^{2} r^{4}}{L R}) \frac{d I}{d t}$ ...(i)

Given, $I = I_{0} \cos (300 t)$

$∴$ $\frac{d I}{d t} = - 300 I_{0} \sin (300 t)$

$M = (\frac{300 π^{2} r^{4}}{L R}) μ_{0} I_{0} \sin (300 t)$

$M = N μ_{0} I_{0} \sin (300 t)$

$N = \frac{300 π^{2} r^{4}}{L R}$

$N = \frac{300 {(\frac{22}{7})}^{2} {(0.1)}^{4}}{(10) (0.005)}$

$N = 5.926 \approx 6$

Q 5 :

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 $ω$ . The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as $α \frac{Q}{2 M} .$ The value of $α$ is _______. [2025]

(1.67)

Magnetic dipole moment $M = I A$

$d M = d I A$

$A = π r^{2} = π {(R \sin θ)}^{2}$

$d I = \frac{d q}{T} = \frac{σ (2 π r) (R d θ) ω}{2 π}$

$d I = \frac{σ 2 π R^{2} ω \sin θ d θ}{2 π}$

$d I = σ R^{2} ω \sin θ d θ$

Magnetic dipole moment:

$M = \int d M = \int_{0}^{π} σ R^{2} ω π R^{2} \sin^{3} θ d θ$

$M = σ R^{4} ω π \int_{0}^{π} \sin^{3} θ d θ$ $(∵ \int_{0}^{π} \sin^{3} θ d θ = \frac{4}{3})$

$M = (\frac{Q}{4 π R^{2}}) R^{4} ω π (\frac{4}{3}) = \frac{Q R^{2} ω}{3}$

Angular momentum, $L = I ω$

$L = (\frac{2}{5} M R^{2}) ω$

$∴$ $\frac{M}{L} = \frac{\frac{Q R^{2} ω}{3}}{\frac{2}{5} M R^{2} ω} = \frac{Q}{2 M} (\frac{5}{3})$

$∴$ $α = \frac{5}{3} = 1.67$

Q 6 :

Two infinitely long straight wires lie in the $x y$ -plane along the lines $x = \pm R .$ The wire located at $x = + R$ carries a constant current $I_{1}$ and the wire located at $x = - R$ carries a constant current $I_{2}$ . A circular loop of radius $R$ is suspended with its centre at $(0, 0, \sqrt{3} R)$ and in a plane parallel to the $x y$ -plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+ \hat{j}$ direction. Which of the following statements regarding the magnetic field $\vec{B}$ is (are) true? [2018]

If $I_{1} = I_{2}$ , then $\vec{B}$ cannot be equal to zero at the origin $(0, 0, 0)$
If $I_{1} > 0$ and $I_{2} < 0$ , then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$
If $I_{1} < 0$ and $I_{2} > 0$ , then $\vec{B}$ can be equal to zero at the origin $(0, 0, 0)$
If $I_{1} = I_{2}$ , then the $z$ -component of the magnetic field at the centre of the loop is $(- \frac{μ_{0} I}{2 R})$

1

2

3

4

Select one or more options

(1, 2, 4)

(1) If $I_{1} = I_{2}$ , then the magnetic fields due to $I_{1}$ and $I_{2}$ at origin $' O'$ will cancel out each other. But the magnetic field at $' O'$ due to the ring will be present. Therefore, $B$ cannot be zero at origin.

(2) If $I_{1} > 0$ and $I_{2} < 0$ , then the magnetic field due to both currents will be in the $+ Z$ direction and add up. The magnetic field due to current $I$ will be in the $- Z$ direction and if its magnitude is equal to the combined magnitudes of $I_{1}$ and $I_{2}$ , then $B$ can be zero at the origin.

(3) If $I_{1} < 0$ and $I_{2} > 0$ , then their net magnetic field at the origin will be in the $- Z$ direction and the magnetic field due to current $I$ at the origin will also be in the $- Z$ direction. Therefore, $\vec{B}$ at origin cannot be zero.

(4) If $I_{1} = I_{2}$ , then the resultant of the magnetic field $B_{R}$ at $P$ is along the $+ X$ direction. Therefore, the magnetic field at $P$ is only due to the current $I$ , which is in
the $- Z$ direction and is equal to $\vec{B} = \frac{μ_{0} I}{2 R} (- \hat{k})$

Q 7 :

Two metallic rings $A$ and $B$ , identical in shape and size but having different resistivities $ρ_{A}$ and $ρ_{B}$ , are kept on top of two identical solenoids as shown in the figure. When current $I$ is switched on in both the solenoids in identical manner, the rings $A$ and $B$ jump to heights $h_{A}$ and $h_{B}$ , respectively, with $h_{A} > h_{B} .$ The possible relation(s) between their resistivities and their masses $m_{A}$ and $m_{B}$ is(are) [2009]

$ρ_{A} > ρ_{B}$ and $m_{A} = m_{B}$
$ρ_{A} < ρ_{B}$ and $m_{A} = m_{B}$
$ρ_{A} > ρ_{B}$ and $m_{A} > m_{B}$
$ρ_{A} < ρ_{B}$ and $m_{A} < m_{B}$

1

2

3

4

Select one or more options

(2, 4)

Induced emf $e = - \frac{d ϕ}{d t}$ . For identical rings, the induced emf will be the same, but the currents will be different.

Given, $h_{A} > h_{B}$

Hence, $v_{A} > v_{B}$ as $(h = \frac{v^{2}}{2 g}) .$

If $ρ_{A} > ρ_{B},$ then $I_{A} < I_{B} .$ In this case, the given condition can be fulfilled if $m_{A} < m_{B} .$

If $ρ_{A} < ρ_{B},$ then $I_{A} > I_{B} .$ In this case, the given condition can be fulfilled if $m_{A} \leq m_{B} .$

Q 8 :

Two wires each carrying a steady current I are shown in four configurations in Column I. Some of the resulting effects are described in Column II. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS. [2007]

Column I Column II

(A)
Point P is situated midway between the wires.

(p) The magnetic fields (B) at P due to the currents in the wires are in the same direction.

(B)
Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii.

(q) The magnetic fields (B) at P due to the currents in the wires are in opposite directions.

(C)
Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii.

(r) There is no magnetic field at P.

(D)
Point P is situated at the common center of the wires.

(s) The wires repel each other.

$A : q, s; B : p; C : q, r; D : q, r$
$A : q, r; B : p; C : q, r; D : q, s$
$A : q, s; B : q, r; C : p; D : q, r$
$A : p; B : q, r; C : q, s; D : q, r$

1

2

3

4

(2)

A : q, r

B at P due to upper wire is in downward direction and due to lower wire is in upward direction. Hence, q is correct.

As P is the mid point, the two magnetic fields cancel out each other. Therefore, r is correct.

B : p

B at P due to current in loop A is along the axial line towards right and due to current in loop B is also along the axial line towards right. Hence B and P due to the currents in the wires are in the same direction.

C : q, r

The magnetic field due to current in loop A at P is equal and opposite to the magnetic field due to current in loop B at P.

D : q, s

'B' at P due to current in inner loop is perpendicular to the plane of paper directed vertically upwards.

'B' at P due to current in outer loop is perpendicular to the plane of paper directed vertically downwards.

As the currents are in opposite directions, the wires repel each other. But net force on each wire is zero.

Q 9 :

Match the following columns: [2006]

Column I Column II

(A) Dielectric ring uniformly charged (p) Constant electrostatic field out of system

(B) Dielectric ring uniformly charged rotating with angular velocity $ω$ (q) Magnetic field strength

(C) Constant current in ring $i$ (r) Electric field (induced)

(D) $i = i_{0} \cos ω t$ (s) Magnetic dipole moment

A – q, s; B – p; C – q, s; D – q, r, s
A – q, s; B – q, r, s ; C – q, s; D – p
A – p; B – q, s; C – q, s; D – q, r, s
A – q, r, s; B – q, s; C – q, s; D – p

1

2

3

4

(3)

(1) Charge on dielectric ring will create an electrostatic field which is time independent.

(2) The rotating charge is like a current. This will create a magnetic field and a magnetic moment.

(3) Constant current in ring, so net charge is zero. Therefore, there will be no time-independent electric field. The current produces a magnetic field and a magnetic moment.

(4) $i = i_{0} \cos ω t$ . A changing magnetic field will be produced. This will create an induced electric field. Also, a changing magnetic moment will be produced.

Q 10 :

Column I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS. [2007]

Column I Column II

(A) A charged capacitor is connected to the ends of the wire. (p) A constant current flows through the wire.

(B) The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion. (q) Thermal energy is generated in the wire.

(C) The wire is placed in a constant electric field that has a direction along the length of the wire. (r) A constant potential difference develops between the ends of the wire.

(D) A battery of constant emf is connected to the ends of the wire. (s) Charges of constant magnitude appear at the ends of the wire.

A : r,s ; B : s ; C : p,q,r ; D : q
A : p,q,r ; B : s ; C : r,s ; D : q
A : p,q,r ; B : r,s ; C : s ; D : q
A : q ; B : r,s ; C : s ; D : p,q,r

1

2

3

4

(4)

A : q

Thermal energy is generated in the wire, when a charged capacitor is connected to the ends of the wire, a variable current (decreasing in magnitude with time) passes through the wire (shown as resistor). The potential difference across the wire also decreases with time. The charge on the capacitor plate also decreases with time.

B : r, s

The wire is moved perpendicular to its length $(ℓ)$ with a constant velocity (v) in a uniform magnetic field (B) perpendicular to the plane of motion $e = B ℓ v$

When $B, ℓ, v$ are constant, e is constant.

A constant potential difference develops across the ends of the wire and charges of constant magnitude appear at the ends of the wire.

C : s

When wire is placed in a constant electric field that has a direction along the length of the wire. The free electrons move under the influence of electric field opposite to the direction of electric field. This movement of $e ⁻$ continues till the electric field inside the wire is zero. Charges of constant magnitude appear at the ends of the wire.

D : p, q, r

Since emf of the battery E, R are constant, a constant current flows in the wire. Due to heating effect of current, thermal energy is generated in the wire. Also a constant potential difference develops between the ends of the wire.

Q 11 :

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 compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$ , with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r$ $(r ≫ a)$ from the center of the loop with its north pole always facing the loop, as shown in the figure below.

The magnitude of magnetic field of a dipole $m$ , at a point on its axis at distance $r$ , is $\frac{μ_{0} m}{2 π r^{3}},$ where $μ_{0}$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_{1}$ and $m_{2}$ , separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_{1} m_{2}}{r^{4}},$ where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

Q. When the dipole $m$ is placed at a distance $r$ from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to [2021]

$\frac{m}{r^{3}}$
$\frac{m^{2}}{r^{2}}$
$\frac{m}{r^{2}}$
$\frac{m^{2}}{r}$

1

2

3

4

(1)

Magnetic flux due to dipole through ring

$ϕ = L i = \frac{μ_{0} m}{2 π r^{3}} \times π a^{2}$ $\Rightarrow i = \frac{μ_{0} m π a^{2}}{2 π r^{3} L}$

or $i \propto \frac{m}{r^{3}}$

$Magnetic moment, m' = i A = π a^{2} i = \frac{μ_{0} m π^{2} a^{4}}{2 π r^{3} L}$

$F = \frac{k m m'}{r^{4}} = \frac{k m^{2} π^{2} a^{4}}{2 π r^{7} L}$

$Therefore work done in bringing the dipole,$

$W = \int F d r \propto \int \frac{m^{2} d r}{r^{7}}$ or, $W \propto \frac{m^{2}}{r^{6}}$

Q 12 :

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 compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$ , with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r$ $(r ≫ a)$ from the center of the loop with its north pole always facing the loop, as shown in the figure below.

The magnitude of magnetic field of a dipole $m$ , at a point on its axis at distance $r$ , is $\frac{μ_{0} m}{2 π r^{3}},$ where $μ_{0}$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_{1}$ and $m_{2}$ , separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_{1} m_{2}}{r^{4}},$ where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

Q. The work done in bringing the dipole from infinity to a distance $r$ from the center of the loop by the given process is proportional to [2021]

$\frac{m}{r^{5}}$
$\frac{m^{2}}{r^{5}}$
$\frac{m^{2}}{r^{6}}$
$\frac{m^{2}}{r^{7}}$

1

2

3

4

(3)

$Magnetic flux due to dipole through ring$

$ϕ = L i = \frac{μ_{0} m}{2 π r^{3}} \times π a^{2}$ $\Rightarrow i = \frac{μ_{0} m π a^{2}}{2 π r^{3} L}$

or $i \propto \frac{m}{r^{3}}$

$Magnetic moment, m' = i A = π a^{2} i = \frac{μ_{0} m π^{2} a^{4}}{2 π r^{3} L}$

$F = \frac{k m m'}{r^{4}} = \frac{k m^{2} π^{2} a^{4}}{2 π r^{7} L}$

$Therefore work done in bringing the dipole,$

$W = \int F d r \propto \int \frac{m^{2} d r}{r^{7}}$ or, $W \propto \frac{m^{2}}{r^{6}}$

Q 13 :

In a thin rectangular metallic strip a constant current $I$ flows along the positive $x$ -direction, as shown in the figure. The length, width and thickness of the strip are $ℓ$ , $w$ and $d$ , respectively.

A uniform magnetic field $\vec{B}$ is applied on the strip along the positive $y$ -direction. Due to this, the charge carriers experience a net deflection along the $z$ -direction. This results in accumulation of charge carriers on the surface $P Q R S$ and appearance of equal and opposite charges on the face opposite to $P Q R S$ . A potential difference along the $z$ -direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

Q. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are $w_{1}$ and $w_{2}$ and thicknesses are $d_{1}$ and $d_{2}$ , respectively. Two points $K$ and $M$ are symmetrically located on the opposite faces parallel to the $x - y$ plane (see figure). $V_{1}$ and $V_{2}$ are the potential differences between $K$ and $M$ in strips 1 and 2, respectively. Then, for a given current $I$ flowing through them in a given magnetic field strength $B$ , the correct statement(s) is(are): [2015]

If $w_{1} = w_{2}$ and $d_{1} = 2 d_{2}$ , then $V_{2} = 2 V_{1}$
If $w_{1} = w_{2}$ and $d_{1} = 2 d_{2}$ , then $V_{2} = V_{1}$
If $w_{1} = 2 w_{2}$ and $d_{1} = d_{2}$ , then $V_{2} = 2 V_{1}$
If $w_{1} = 2 w_{2}$ and $d_{1} = d_{2}$ , then $V_{2} = V_{1}$

1

2

3

4

Select one or more options

(1, 4)

When magnetic force balances electric force

$F_{B} = F_{E}$ $\Rightarrow q v_{d} B = q E$

$∴$ $v_{d} B = \frac{V}{w}$ $[∵ V = E \times w]$

$∴$ $V = w v_{d} B = w [\frac{I}{n e w d}] \times B$ $[v_{d} = \frac{I}{n e A} = \frac{I}{n e w d}]$

$∴$ $V = \frac{I}{n e d} \times B$

or, $V \propto \frac{1}{d}$ $\Rightarrow V_{1} d_{1} = V_{2} d_{2}$

If $d_{1} = 2 d_{2},$ $V_{2} = 2 V_{1}$

and if $d_{1} = d_{2}$ , $V_{2} = V_{1}$

Q 14 :

In a thin rectangular metallic strip a constant current $I$ flows along the positive $x$ -direction, as shown in the figure. The length, width and thickness of the strip are $ℓ$ , $w$ and $d$ , respectively.

A uniform magnetic field $\vec{B}$ is applied on the strip along the positive $y$ -direction. Due to this, the charge carriers experience a net deflection along the $z$ -direction. This results in accumulation of charge carriers on the surface $P Q R S$ and appearance of equal and opposite charges on the face opposite to $P Q R S$ . A potential difference along the $z$ -direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons.

Q. Consider two different metallic strips (1 and 2) of same dimensions (length $l$ , width $w$ and thickness $d$ ) with carrier densities $n_{1}$ and $n_{2}$ , respectively. Strip 1 is placed in magnetic field $B_{1}$ and strip 2 is placed in magnetic field $B_{2}$ , both along positive $y$ -directions. Then $V_{1}$ and $V_{2}$ are the potential differences developed between $K$ and $M$ in strips 1 and 2, respectively. Assuming that the current $I$ is the same for both the strips, the correct option(s) is(are) [2015]

If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ , then $V_{2} = 2 V_{1}$
If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ , then $V_{2} = V_{1}$
If $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ , then $V_{2} = 0.5 V_{1}$
If $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ , then $V_{2} = V_{1}$

1

2

3

4

Select one or more options

(1, 3)

$∵$ $V = \frac{I}{n e d} \times B$

$∴$ $V \propto \frac{B}{n}$ $\Rightarrow \frac{V_{1} n_{1}}{B_{1}} = \frac{V_{2} n_{2}}{B_{2}}$

If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ $\Rightarrow V_{2} = 2 V_{1}$

and if $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ $\Rightarrow V_{2} = 0.5 V_{1}$

Q 15 :

The figure shows a circular loop of radius $a$ with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is $d$ . The loop and the wires are carrying the same current $I$ . The current in the loop is in the counterclockwise direction if seen from above.

Q. When $d \approx a$ but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height $h$ above the loop. In that case [2014]

current in wire 1 and wire 2 in the direction PQ and RS, respectively and $h \approx a$
current in wire 1 and wire 2 in the direction PQ and SR, respectively and $h \approx a$
current in wire 1 and wire 2 in the direction PQ and SR, respectively and $h \approx 1.2 a$
current in wire 1 and wire 2 in the direction PQ and RS, respectively and $h \approx 1.2 a$

1

2

3

4

(3)

$Here B_{1} \sin θ and B_{2} \sin θ cancelled each other.$

For zero magnetic field at $' P'$

Magnetic field due to current carrying circular loop = Magnetic field due to straight wires

$B = B_{1} \cos θ + B_{2} \cos θ = 2 B_{1} \cos θ$

$\frac{μ_{0} I a^{2}}{2 {(a^{2} + h^{2})}^{3 / 2}} = 2 [\frac{μ_{0} I}{2 π \sqrt{a^{2} + h^{2}}}] \times \frac{a}{\sqrt{a^{2} + h^{2}}}$

Solving we get,

$h \approx 1.2 a$

The current is from P to Q and R to S in wire 1 and wire 2 respectively.

Q 16 :

The figure shows a circular loop of radius $a$ with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is $d$ . The loop and the wires are carrying the same current $I$ . The current in the loop is in the counterclockwise direction if seen from above.

Q. Consider $d ≫ a$ , and the loop is rotated about its diameter parallel to the wires by $30 °$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop). [2014]

$\frac{μ_{0} I^{2} a^{2}}{d}$
$\frac{μ_{0} I^{2} a^{2}}{2 d}$
$\frac{\sqrt{3} μ_{0} I^{2} a^{2}}{d}$
$\frac{\sqrt{3} μ_{0} I^{2} a^{2}}{2 d}$

1

2

3

4

(2)

We know torque

$\vec{τ} = \vec{M} \times \vec{B} = M B \sin θ$

$= (I \times π a^{2}) \times [2 \times \frac{μ_{0} I}{2 π d}] \sin 30 °$

$∴$ $τ = \frac{μ_{0} I^{2} a^{2}}{2 d}$

Q 17 :

Advanced countries are making use of powerful electromagnets to move trains at very high speed. These trains are called maglev trains (abbreviated from magnetic levitation). These trains float on a guideway and do not run on steel rail tracks.

Instead of using an engine based on fossil fuels, they make use of magnetic field forces. The magnetized coils are arranged in the guideway which repels the strong magnets placed in the train's under carriage. This helps train move over the guideway, a technic called electro-dynamic suspension. When current passes in the coils of guideway, a typical magnetic field is set up between the undercarriage of train and guideway which pushes and pull the train along the guideway depending on the requirement.

The lack of friction and its aerodynamic style allows the train to move at very high speed.

Q. The levitation of the train is due to [2006]

Mechanical force
Electrostatic attraction
Electrostatic repulsion
Magnetic repulsion

1

2

3

4

(4)

The levitation of the train is due to magnetic repulsion. The magnetised coils running along the track repel large magnets on the train's under carriage.

Q 18 :

Advanced countries are making use of powerful electromagnets to move trains at very high speed. These trains are called maglev trains (abbreviated from magnetic levitation). These trains float on a guideway and do not run on steel rail tracks. Instead of using an engine based on fossil fuels, they make use of magnetic field forces.

The magnetized coils are arranged in the guideway which repels the strong magnets placed in the train's under carriage. This helps train move over the guideway, a technic called electro-dynamic suspension. When current passes in the coils of guideway, a typical magnetic field is set up between the undercarriage of train and guideway which pushes and pull the train along the guideway depending on the requirement.

The lack of friction and its aerodynamic style allows the train to move at very high speed.

Q. The disadvantage of maglev trains is that [2006]

More friction
Less pollution
Less wear & tear
High initial cost

1

2

3

4

(4)

High initial cost.

Q 19 :

Advanced countries are making use of powerful electromagnets to move trains at very high speed. These trains are called maglev trains (abbreviated from magnetic levitation). These trains float on a guideway and do not run on steel rail tracks. Instead of using an engine based on fossil fuels, they make use of magnetic field forces.

The magnetized coils are arranged in the guideway which repels the strong magnets placed in the train's under carriage. This helps train move over the guideway, a technic called electro-dynamic suspension. When current passes in the coils of guideway, a typical magnetic field is set up between the undercarriage of train and guideway which pushes and pull the train along the guideway depending on the requirement.

The lack of friction and its aerodynamic style allows the train to move at very high speed.

Q. The force which makes maglev move [2006]

Gravitational field
Magnetic field
Nuclear forces
Air drag

1

2

3

4

(2)

Maglev is the abbreviation of magnetic levitation. The magnetic force will pull the maglev trains.

Topic Question Set

Want Us to Email you About Special Offers & Updates?

	Column I		Column II
(A)	Point P is situated midway between the wires.	(p)	The magnetic fields (B) at P due to the currents in the wires are in the same direction.
(B)	Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii.	(q)	The magnetic fields (B) at P due to the currents in the wires are in opposite directions.
(C)	Point P is situated at the mid-point of the line joining the centers of the circular wires, which have same radii.	(r)	There is no magnetic field at P.
(D)	Point P is situated at the common center of the wires.	(s)	The wires repel each other.

	Column I		Column II
(A)	Dielectric ring uniformly charged	(p)	Constant electrostatic field out of system
(B)	Dielectric ring uniformly charged rotating with angular velocity $ω$	(q)	Magnetic field strength
(C)	Constant current in ring $i$	(r)	Electric field (induced)
(D)	$i = i_{0} \cos ω t$	(s)	Magnetic dipole moment

	Column I		Column II
(A)	A charged capacitor is connected to the ends of the wire.	(p)	A constant current flows through the wire.
(B)	The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion.	(q)	Thermal energy is generated in the wire.
(C)	The wire is placed in a constant electric field that has a direction along the length of the wire.	(r)	A constant potential difference develops between the ends of the wire.
(D)	A battery of constant emf is connected to the ends of the wire.	(s)	Charges of constant magnitude appear at the ends of the wire.