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CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 11 - MCQExams.com
CBSE
Class 12 Medical Physics
Moving Charges And Magnetism
Quiz 11
A charged particle moves through a magnetic field perpendicular to its direction, then
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the momentum changes but the K.E is constant
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both momentum and K.E of the particle are not constant
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both momentum and K.E of the particle are constant
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K.E changes but the momentum is constant
Explanation
Momentum changes but K.E remains constant because particles gets deflected and moved in a curved path so the velocity changes as the direction of the speed changes hence the movement on the body also changes while K.E unchanged.
Two wires $$A$$ and $$B$$ are carrying currents $$I_{1}$$ and $$I_{2}$$ as shown in the figure. The separation between them is $$d$$. A third
wire $$C$$ carrying a current $$I$$ is to be kept parallel to them at a distance $$x$$ from $$A$$ such that the net force acting on it is
zero. The possible values of $$x$$
are
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$$x=\dfrac{I_{1}}{I_{1}-I_{2}}$$d and $$x=\dfrac{I_{2}}{I_{1}-I_{2}}d$$
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$$x=\pm(\dfrac{I_{1}d}{I_{1}-I_{2}})$$
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$$x=\dfrac{I_{1}}{I_{1}+I_{2}}d$$ and $$x=\dfrac{I_{2}}{I_{1}-I_{2}}d$$
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$$x=\dfrac{I_{2}}{I_{1}+I_{2}}d$$ and $$x=\dfrac{I_{1}}{I_{1}-I_{2}}d$$
Explanation
Net force on wire carrying current $$I$$ per unit length is
$$\dfrac{\mu_0 I_1I}{2\pi x}+\dfrac{\mu_0I_2I}{2\pi(d-x)}=0$$
$$\dfrac{I_1}{x}=\dfrac{I_2}{x-d}$$
$$\Rightarrow x=\dfrac{I_1d}{I_1-I_2}$$
An electron, a proton and a $$He^+$$ ion projected into a magnetic field with same kinetic energy, with velocities being perpendicular to the magnetic field. The order of the radii of cirlces traced by them is:
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$$r_p > r_{He^+} > r_e$$
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$$r_{He^+} > r_p > r_e$$
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$$r_p = r_{He^+} > r_e$$
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None of these
Explanation
radius of circle is given by
$$r = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2 mk}}{qB} = \dfrac{\sqrt{2m}}{qB} \sqrt{k}$$
where K is kinetic energy
For poor
$$r_p = \dfrac{\sqrt{2m_p}}{eB} \sqrt{k}$$
for electron $$r_e = \dfrac{\sqrt{2m_e}}{eB} \sqrt{K}$$
for $$He^+ \, r_{He^+} = \dfrac{\sqrt{2 \times 4 m_p}}{eB} \sqrt{K} = \dfrac{\sqrt[2]{2m_p}}{eB} \sqrt{K}$$
Clearly $$r_{He^+} > r_p > r_e$$
A long straight wire carries a current of $$50\, A$$. An electron moving at $$10^7\, m/s$$ is $$5\, cm$$ away from the wire. The force acting on electron if its velocity is directed towards the wire will be :
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$$1.6\times 10^{-6}\, N$$
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$$3.2\times 10^{-16}\, N$$
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$$4.8\times 10^{-16}\, N$$
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$$1.8\times 10^{-16}\, N$$
Explanation
Given:
Current in wire= 50A
Speed of electron=$$10^{7}ms^{-1}$$
Solution:
Let us assume the current is directed in +y direction.
As we know, magnetic field due to straight wire is given by,
$$B=\frac{\mu _{0}i}{2\pi r}$$
$$=\frac{4\pi \times 10^{-7}\times 50}{2\pi \times 0.05}$$
$$=0.0002T$$
According to question, v and B are perpendicular to each other.
$$F=qvb$$
Substituting the values we get,
$$=1.6\times 10^{-19}\times 10^{7}\times 0.0002$$
$$=3.2\times 10^{-16}N$$
Hence, the correct option is (B).
Magnetization for vacuum is......
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Negative
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Positive
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Infinite
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zero
Explanation
Solution :
Magnetization for vacuum
is the measure of the corresponding response of a material to an electric field in electrostatics. Which
means it attracts the electrons. soo this magnetization is positive.
The Correct Opt : B
A particle of charge $$ 1 \mu C $$ is at rest in a magnetic field $$ \overrightarrow {B} = -2 \overrightarrow {k} $$ tesla,Magnetic Lorentz force on the charge particle with respect to an observer moving with velocity $$ \overrightarrow {v} = -5 \hat {i} ms^{-1} $$ will be
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$$ +10^{-5} \hat {j} N $$
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zero
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$$ -10^{-5} \hat {j} N $$
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$$ -10^{-6} \hat {j} N $$
Explanation
Given
Charge= $$10^{-6}$$C
B= -2T $$ \hat{k}$$
v= -5m/s $$\hat{i}$$
Solution
Lorentz forve is reference dependent therefore for moving reference velocity of object is in opposite direction that of moving reference
Therefore velocity of charge= +5m/s $$\hat{i}$$
F=q(vxB)
$$F=10^{-6}(5*2)(\hat{i} * \hat{k})$$
F=10^{-5} $$\hat{j}$$
The correct option is D
In moving coil galvanometer, strong horses shoe magnet of concave shaped pole pieces is used to?
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Increase space for rotation of coil
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Reduce weight of galvanometer
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Protect magnetic field which is parallel to plane of coil at any position
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Make magnetic induction weak at the cnetre
Explanation
The moving coil of a moving coil galvanometer moves in a magnetic field produced by a permanent magnet. When a current passes through the coil, its sides which are perpendicular to the magnetic field, experience equal and opposite force. This force is separated by the width of the coil. These two equal and opposite forces separated by the width of the coil constitute a couple, which rotates the coil.
We want the current-carrying coil to be always perpendicular to the magnetic field, even when it has rotated. The magnetic field produced by concave-shaped pole faces is always radial. A radial field is always perpendicular to a conductor rotating about an axis passing through the centre of the concave-shaped pole faces and parallel to the faces.
Pole faces are concave to make the magnetic field radial to keep it always normal to the moving coil.
A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in
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end-on position
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broadside-on position
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both
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none
Explanation
The correct answer should be option a.
The magnetic moment of the current carrying loop shown in the figure is equal is :
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$$\dfrac {I (b^2 + 2ab)\theta}{2}$$
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$$I a b \theta$$
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$$\dfrac {I (a^2 + ab)\theta}{2}$$
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$$none\ of\ the\ above$$
The magnitude of magnetic moment of the current loop in the figure is :
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$$I a^2$$
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$$\sqrt {2} I a^2$$
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$$zero$$
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$$none\ of\ the\ above$$
If an electron describes half a revolution in a circle of radius $$r$$ in a magnetic field $$B$$, the energy acquire by it is :
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$$zero$$
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$$\dfrac {1}{2}mv^2$$
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$$\dfrac {1}{2}\left (\dfrac {1}{2}mv^2 \right)$$
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$$\pi r (Bev)$$
Explanation
Explanation:
As per rule, the magnetic force is only responsible for changing the direction of electrons while perpendicular to velocity of electrons.
So, there would be no work done as magnetic force can not accelerate or decelerate electrons.
"Hence, energy acquired by electron will be zero"
A wire of length $$L\ m$$ carrying a current $$I$$ amp is bent in the from of a circle. The magnitude of magnetic moment is :
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$$\dfrac {IL^2}{4\pi}$$
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$$\dfrac {IL^2}{2\pi}$$
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$$\dfrac {IL}{4\pi}$$
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$$I\pi L^2$$
Two parallel wires carrying equal currents in opposite direction are placed at $$ x = \pm a $$ parallel to y -axis with z =0 . Magnetic field at origin O is $$ B_1 $$ and at P ( 2a, 0 , 0) is $$ B_2 $$ then , the ratio $$ B_1 /B_2 $$is
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3
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1/2
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1/3
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2
Explanation
As shown in fig.
Magnetic field at Point O due to both wire, $$ \overrightarrow {B_1} = -2 ( \dfrac { \mu_0}{ 2\pi} \dfrac {i}{a} ) \hat {k} = \dfrac { \mu_0 i}{ \pi a } \hat {k} $$
Magnetic field at Point P due to both wire
$$ \overrightarrow {B_2} = \dfrac { \mu_0}{ 2 \pi } \dfrac {i}{a } \hat {k} - \dfrac { \mu_0 }{2 \pi} \dfrac {i}{ 3a} \hat {k} = \dfrac { \mu_0}{3 \pi } \dfrac {i}{a} \hat {k} $$
$$ \therefore B_1 /B_2 = 3 $$
Find $$ \oint \vec{B}\cdot \vec{dl} $$ over following (direction in which integration has to be performed is indicated by arrows
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Two infinitely long linear conductors are arranged perpendicular to each other and are in mutually perpendicular planes as shown in fig. if $$ I_1 = 2A $$ along the y-axis , $$ I_2 = 3A $$ along -ve z-axis and AP = AB = 1 cm , the value of magnetic field strength field strength $$ \overrightarrow {B} $$ at P is
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$$ ( 3 \times 10^{-5} T ) \hat {j} + ( -4 \times 10^{-5} T ) \hat {k} $$
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$$ ( 3 \times 10^{-5} T ) \hat {j} + ( 4 \times 10^{-5} T ) \hat {k} $$
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$$ ( 4 \times 10^{-5} T ) \hat {j} + ( 3 \times 10^{-5} T ) \hat {k} $$
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$$ ( -3 \times 10^{-5} T ) \hat {j} + ( 4 \times 10^{-5} T ) \hat {k} $$
Explanation
Magnetic field strength at P due to $$ I_1 $$
$$ \overrightarrow {B}_1 = \frac { \mu_0 I_1}{ 2 \pi (AP) } \hat {k } = \frac { 4 \pi \times 10^{-7} \times 2 }{ 2 \pi \times 1 \times 10^{-2} } \hat {k} = ( 4 \times 10^{-5} T) \hat {j} $$
hence , $$ \overrightarrow {B} = ( 3 \times 10^{-5} T ) \hat {j} + ( 4 \times 10^{-5} T ) \hat {k} $$
Ratio of the currents $$ I_1 $$ and $$ I_2 $$ flowing through the circular and straight parts is
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$$ \frac { \sqrt {3} }{ 2 \pi} $$
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$$ \frac { 2 \sqrt {3} }{ \pi} $$
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$$ \frac { 3 \sqrt {3} }{ 2 \pi } $$
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$$ \frac { 3 \sqrt {3} }{ 2 \sqrt {2} \pi } $$
Explanation
IF $$ I_1 $$ and$$ I_2 $$ be the currents in circular and straight parts, respectively , and $$ B_1 $$ and $$ B_2 $$ the magnetic field due to them, The direction of $$B_1$$ and $$B_2$$ are opposite to each other according to right hand thumb rule.
Now, $$B_1$$= $$\frac {\mu_0 I_1}{2R} \times \frac {240}{360}$$ [Here part of loop is present ]
$$ B_1 = \frac { \mu_0 I_1}{ 2R} \times \frac {2}{3} = \frac { \mu_0 I_1 }{3R} $$
$$B_2 = \frac { \mu_0 I_2}{ 4 \pi [ R cos 60^0] } [ 2 sin 60^0 ]= \frac { \sqrt {3} \mu_0 I_2 }{ 2\pi R}$$
For the total field at 'O; to be zero ,$$ \frac { \mu_0 I_1}{ 3R } = \frac {\sqrt {3} \mu_0 I_2 }{ 2 \pi R } \Rightarrow \frac { I_1}{I_2} = \frac { 3 \sqrt {3} }{ 2 \pi } $$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
Explanation
Work done by the magnetic field is always zero.
The magnetic force $$q(V\times B)$$
is normal to $$V$$
and work done by it is zero.
Hence Statement 1 is false.
If force acts opposite to the direction of velocity its work done is negative
Two very thin metallic wires placed along X and Y -axes carry equal currents as shown in fig. Ab and CD are lines at $$ 45^0 $$ with the axes . the magnetic fields will be zero on the line
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AB
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CD
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segment OB only of line AB
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segment OC only of line CD
Explanation
All points on AB are equidistant from the two wires . so magnetic fields are equal and opposite . thus, they cancel out.
Two long thin wires ABC and DEF are arranged as shown in fig. they carry equal current I as shown. the magnitude of the magnetic field at O is
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zero
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$$ \mu_0I / 4 \pi a $$
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$$ \mu_0 I / 2 \pi a $$
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$$ \mu_0 I / 2 \sqrt {2} \pi a $$
Explanation
AB and DE do not give magnetic field at O.
Magnetic field at O due to $$ BC, B_1= \dfrac { \mu_0 I}{ 4 \pi a} $$
it is directed 'up '
Magnetic field at O due to $$EF, B_2 = \dfrac { \mu_0 I }{ 4 \pi a } $$ it is also directed up.
So,
$$ B=B_1+B_2 = \dfrac { \mu_0 I}{ 2 \pi a} $$
Bolt-savart law indicates that the moving electrons (velocity v) produce a magnetic field b such that
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.$$B \bot v$$
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B $$B || v.$$
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it obeys inverse cube law.
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it is along the line joining the electron and point of observation.
Explanation
By Biot-savart law, $$ dB=\dfrac{I.dlsin\theta }{r^{2}} $$
Or $$ dB= \dfrac{I\times dl}{r} $$
acc to Biot-Savart law , if magnetic field is not perpendicular to the motion of charge then it will not move in helical path, which is not possible for motion of a charge in magnetic field.
So the magnetic field is perpendicular to the direction of flow of charge verifies answer 'a'.
A Toroid of turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m
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is non-zero and points in the z-direction by symmetry.
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point along the axis of the toroid
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is zero. otherwise there would be a field falling as $$ \dfrac{1}{r^{3}} $$ at large distances outside the toroid.
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is pointing readily outwards.
Explanation
We know that there is no magnetic field outside the toroid.
So outside, the M.F. falls very rapidly as it is inversely proportional to third power of distance from center of toroid.
By Amperian circuital law, there is no any current so no magnetic moment outside.
Verifies answer (c).
A proton (mass $$=1.67\times 10^{-27} kg$$ and charge $$=1.6\times 10^{-19}C$$) enters perpendicular to a magnetic field of intensity $$2$$ weber $$/m^2$$ with a velocity $$3.4\times 10^7\ m/sec$$. The acceleration of the proton should be
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$$6.5\times 10^{15}m/sec^2$$
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$$6.5\times 10^{13}m/sec^2$$
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$$6.5\times 10^{11}m/sec^2$$
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$$6.5\times 10^{9}m/sec^2$$
Explanation
The force experienced by the proton due to the magnetic field provides an acceleration to it.
$$F=ma=qvB $$
$$\Rightarrow a=\dfrac {qvB}{m}$$
$$=\dfrac {1.6\times 10^{-19}\times 2\times 3.4\times 10^{7}}{1.67\times 10^{-27}}$$
$$=6.5\times 10^{15}\ m/sec$$
Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert
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An attractive force on each other
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An repulsive force on each other
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No force on each other
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A rotational torque on each other
Explanation
Two straight conductors carry current in same direction, then attractive force acts between them.
The radius of curvature of the path of the charges particles in a uniform magnetic field is directly proportional to
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The charge on the particle
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The momentum on the particle
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The energy on the particle
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The intensity of the field
Explanation
The radius of the circular path in the magnetic field is given by:
$$r=\dfrac{mv}{qB}$$
We know that $$mv$$ is the momentum of the body.
$$\therefore r=\dfrac {p}{qB}$$
$$\Rightarrow r\propto p$$
Assertion : If an electron is not deflected while passing through a certain region of space, then only possibility is
that there is no magnetic region.
Reason : Force is directly proportional to the magnetic field applied.
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If both assertion and reason are true and the reason is the correct explanation of the assertion.
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If both assertion and reason are true but reason is not the correct explanation of the assertion.
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If assertion is true but reason is false.
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If the assertion and reason both are false.
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If assertion is false but reason is true.
Explanation
In this case we can not be sure about the absence of the magnetic field because if the electron moving parallel to the direction of magnetic field, the angle between velocity and applied magnetic field is zero $$(F = 0)$$. Then also electron passes without deflection. Also $$F=evB \sin\theta\Rightarrow F\propto B.$$
Two long parallel wires $$P$$ and $$Q$$ are both perpendicular to the plane of the paper with distance $$5\ m$$ between them. If $$P$$ and $$Q$$ carry current of $$2.5\ amp$$ respectively in the same direction, then the magnetic field at a point half way between the wires is
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$$\dfrac {\sqrt 3\mu_0}{2\pi}$$
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$$\dfrac {\mu_0}{\pi}$$
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$$\dfrac { 3\mu_0}{2\pi}$$
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$$\dfrac {\mu_0}{2\pi}$$
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$$\dfrac {\sqrt 3\mu_0}{\pi}$$
Explanation
In the following figure magnetic field at mid point $$M$$ is given by
$$B_{Net}=B_Q -B_P$$
$$=\dfrac{\mu_0}{4\pi}.\dfrac 2r ( i_Q -i_P)$$
$$=\dfrac {\mu_0}{4\pi}\times \dfrac {2}{2.5}(5-2.5)=\dfrac {\mu_0}{2\pi}$$
Magnetic dipole moment is a
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Scalar quantity
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Vector quantity
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Constant quantity
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None of these
Explanation
Magnetic dipole moment is a Vector quantity.
A beam of protons and $$\alpha$$-particles are successively accelerated in a cyclotron. The ratio of the normal magnetic field to be applied to the cyclotron so that protons and $$\alpha$$-particles have the same period of rotation is
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$$1 : 4$$
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$$4 : 1$$
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$$1 : 2$$
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$$2 : 1$$
Explanation
For cyclotron,
Time period of revolution is given as:
$$T=\dfrac{2\pi m}{Bq}$$
Where, m = mass, B = magnetic field intensity and q = charge.
Now, let mass of a proton is 'm' and charge is 'q'.
Then, $$T_p=\dfrac{2\pi m}{B_pq}$$ ......(i) (For Proton)
Now, for alpha particle, $$(He^{2+})$$
$$m_{He}=4\times m_p=4m$$
$$q_{He}=2q_p=2q$$
Now,
$$T_{\alpha}=\dfrac{2\pi(4m)}{B_{\alpha}(2q)}$$
$$\dfrac{B_p}{B_{\alpha}}=\dfrac12$$
Hence, normal magnetioc field for same time period protons and $$\alpha-$$particle is in ratio $$1:2$$.
Two long and parallel wires are at a distance of $$0.1\ m$$ and a current of $$5\ A$$ is flowing in each of these wires. The force per unit length due to these wires will be
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$$5 \times 10^{-5}\ N/m$$
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$$5 \times 10^{-3}\ N/m$$
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$$2.5 \times 10^{-5}\ N/m$$
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$$2.5 \times 10^{-4}\ N/m$$
Explanation
step-1: given data
A magnetic field
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Always exerts a force on a charged particle
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Never exerts a force on a charged particle
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Exerts a force, if the charges particle is moving across the magnitude field lines
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Exerts a force, if the charges particle is moving along the magnitude field lines
Explanation
The force exerted by the magnetic field is given by:
$$\vec F=q\vec v \times \vec B$$
So, the force is exerted only when the charge particle is moving across the magnetic field lines.
The pole piece of the magnet used in a pivoted coil galvanometer are
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Plane surface of a bar magnet
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Plane surface of a horse-shoe magnet
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Cylindrical surface of a bar magnet
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Cylindrical surface of a horse-shoe magnet
Explanation
In a suspended coil galvanometer the strength of the current is not proportional to the deflectionϕ. Hence, the current cannot be read directly. To remove this difficulty, it is necessary that in every position of the coil, the plane of the coil be parallel to the magnetic field. It is for this purpose the coil is suspended between cylindrically cut pole pieces and a soft iron core is placed in the coil without touching it anywhere.
Lorentx force can be calculate dby using the formula
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$$\vec F=q(\vec E+\vec v\times \vec B)$$
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$$\vec F=q(\vec E-\vec v\times \vec B)$$
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$$\vec F=q(\vec E+\vec v . \vec B)$$
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$$\vec F=q(\vec E\times \vec v+\vec B)$$
Explanation
Lorentz force is given by
$$\vec F=\vec {F_e}+\vec {F_m}$$
$$=q\vec E+q(\vec v \times \vec B)$$
$$=q [\vec E+(\vec v\times \vec B)]$$
A long wire $$AB$$ is placed on a table. Another wire $$PQ$$ of mass $$1.0\ g$$ and length $$50\ cm$$ is set to slide on two rails $$PS$$ and $$QR$$. A current of $$50\ A$$ is passed through the wires. At what distance above $$AB$$, will the wire $$PQ$$ be in equilibrium
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$$25\ mm$$
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$$50\ mm$$
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$$75\ mm$$
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$$100\ mm$$
Explanation
Suppose in equilibrium wire $$PQ$$ lies at a distance $$r$$ above the wire $$AB$$
Hence in equilibrium
$$mg=Bil$$
$$\Rightarrow mg=\dfrac{\mu_{0}}{4\pi}\left(\dfrac{2i}{r}\right)\times il$$
$$\Rightarrow 10^{-3}\times 10=10^{-7}\times \dfrac{2\times (50)^{2}}{r}\times 0.5$$
$$\Rightarrow r=25\ mm$$
If a proton, deutron and $$\alpha-$$particle or being accelerated by the same potential difference enters perpendicular to the magnetic field, then the ratio of their kinetic energy is
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$$1:2:2$$
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$$2:2:1$$
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$$1:2:1$$
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$$1:1:2$$
Explanation
Kinetic energy is magnetic field remains constant ant it is
$$K=q\ V$$
$$\Rightarrow K\propto q (V= constant)$$
$$\therefore K_p: K_d : K_\alpha =q_p :q_d : q_\alpha =1:1:2$$
Two parallel conductors $$A$$ and $$B$$ equal length carry currents $$I$$ and $$10\ I$$, respectively, in the same direction. Then
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$$A$$ and $$B$$ repel each other with same face.
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$$A$$ and $$B$$ attract each other with same face.
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$$A$$ will attract $$B$$, but $$B$$ will repel $$A$$
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$$A$$ and $$B$$ attract each other with difference face.
Explanation
By Fleming left hand rule. we can see that if current flow is same direction then their Force direct to each other i.e they will attract each other.
A current loop placed in a magnetic field behaves like a
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Magnetic dipole
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Magnetic substance
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Magnetic pole
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All are true
Explanation
A current loop placed in a magnetic field can behaves like a magnetic dipole.
Two long straight wires are set parallel to each other. Each carries a current $$i$$ in the same direction and the separation between them is $$2r$$. The intensity of the magnetic field midway between them is
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$$\mu_0 i/r$$
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$$4\mu_0i/r$$
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$$Zero$$
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$$\mu_0i/4r$$
Explanation
At $$P$$
$$\Rightarrow B_{net}=B_1-B_2$$
Since $$| B_1 |=| B_2|$$
So $$B_{net}=0$$
In case Hall effect for a strip having charge $$Q$$ and area of cross-section $$A$$, the Lorentz force is
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Directly proportional to $$Q$$
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Inversely proportional to $$Q$$
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Inversely proportional to $$A$$
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Directly proportional to $$A$$
Explanation
$$\vec F = Q[\vec E + (\vec v \times \vec B)]$$
So, $$F \propto Q$$
" On flowing current in a conducting wire the magnetic field produces around it". It is a law of
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Lenz
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Ampere
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Ohm
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Maxwell
Explanation
The aboove statement is ampere's law.
Which of the following statement is true.
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The presence of a large magnetic flux through a coil maintains a current in the coil if the circuit is continuous
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A coil of a metal wire kept stationary in a uniform magnetic field an e.m.f include in it
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A charged particle enters a region of uniform magnetic field at an angle $$85^o$$ to the magnetic lines of force, the path of the particles is a circle
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There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it
Explanation
When charged particle enters perpendicularly in a magnetic field,it moves on a circular path with a constant speed. Hence it's kinetic energy also remains constant.
An electron is moving on a circular path of radius $$r$$ with speed $$v$$ in a transverse magnetic field $$B$$. $$e/m$$ for it will be
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$$\dfrac {v}{rB}$$
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$$\dfrac {B}{rv}$$
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$$Bvr$$
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$$\dfrac {vr}{B}$$
Explanation
The radius is given by $$r=\dfrac {mv}{eB}\Rightarrow \dfrac {e}{m}=\dfrac {v}{rB}$$
When a magnetic field is applied in a direction perpendicular to the direction of cathode rays, then their
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Energy increase
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Energy decrease
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Momentum increase
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Momentum and energy remain unchanged
Explanation
Since force is perpendicular to direction of motion, energy and magnitude of momentum remains constant.
A small cylindrical soft iron piece is kept in a galvanometer so that
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A radial uniform magnetic fields is produced
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A uniform magnetic fields is produced
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There is a steady deflection of the coil
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All of these
Explanation
A soft iron core is used in a moving coil galvanometer. The soft iron core attracts the magnetic lines of force and hence the strength of the magnetic field increases if we use soft iron core. Thus the sensitivity of galvanometer increases. Also the use of soft iron core makes the magnetic field radial (i.e the plane of the coil will be always parallel to the direction of magnetic field).
Mixed $$He^+$$ and $$O^{2+}$$ ions (mass of $$He^+=4$$ amu and that of $$O^{2+}=16$$ amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then
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$$He^+$$ ions will be deflected more than those of $$O^{2+}$$
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$$He^+$$ ions will be deflected less than those of $$O^{2+}$$
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All the ions will be deflected equally
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No ions will be deflected
Explanation
$$r=\dfrac {\sqrt {2mK}}{qB}\Rightarrow r\propto \dfrac {\sqrt m}{q}\Rightarrow \dfrac {r_{He^+}}{r_{O^{++}}}=\sqrt {\dfrac {m_{He^+}}{m_{O^{++}}}}\times \dfrac {q_{O^{++}}}{q_{He^+}}$$
$$=\sqrt {\dfrac {4}{16}}\times \dfrac {2}{1}=\dfrac {1}{1}$$. Then will deflect equality.
A beam of electrons and protons move parallel to each other in the same direction, then they
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Attract each other
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repel each other
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No relation
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Neither attract nor repel
Explanation
The flow of positive chage is taken as the direction of current.
So, here the currents are in opposite direction so, they will repel each other.
The magnetic potential at a point on the axial line of a bar magnet of dipole moment $$ M $$ is $$ V. $$ What is the magnetic potential due to a bar magnet of dipole moment $$ \frac{M}{4} $$ at the same point
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$$ 4 V $$
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$$ 2 V $$
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$$ \frac{V}{2} $$
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$$ \frac{V}{4} $$
Explanation
Magnetic potential at a distance $$ d $$ from the bar magnet on it's axial line is given by
$$ V=\dfrac{\mu_0}{4 \pi}. \dfrac{M}{d^2} \Rightarrow V \propto M \Rightarrow \dfrac{V_1}{V_2}=\dfrac{M_1}{M_2} $$
$$ \Rightarrow \dfrac{V}{V_2}=\dfrac{M}{M/4} \Rightarrow V_2 = \dfrac{V}{4} $$
Classify each of the following statements as a characteristic.
The force exerted on a charged object is proportional to its speed.
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of electric forces only
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of magnetic forces only,
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of both electric and magnetic forces, or
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of neither electric nor magnetic forces.
Explanation
$$F_B=|q|vB\sin\theta$$ is non-zero unless $$\theta=\pm90^{o}$$.
Classify each of the following statements as a characteristic.
The force exerted on a moving charged object is zero.
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of electric forces only
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of magnetic forces only,
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of both electric and magnetic forces, or
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of neither electric nor magnetic forces.
Explanation
But $$F_B=|q|vB\sin\theta$$ is zero if $$\theta=\pm90^{o}$$
A long, vertical, metallic wire carries downward electric current.
What would be the direction of the field if the current consisted of positive charges moving downward instead of electrons moving upward?
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north
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south
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east
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west
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up
Explanation
The direction of the magnetic field at a given point is determined by the direction of the conventional current that creates it.
In Figure, assume $$I_1=2.00 \,A$$ and $$I_2=6.00 \,A$$. What is the relationship between the magnitude $$F_1$$ of the force exerted on wire $$1$$ and the magnitude $$F_2$$ of the force exerted on wire $$2$$?
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$$F_1=6F_2$$
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$$F_1=3F_2$$
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$$F_1=F_2$$
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$$F_1=\dfrac{1}{3}F_2$$
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$$F_1=\dfrac{1}{6}F_2$$
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Practice Class 12 Medical Physics Quiz Questions and Answers
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