Magnetic effects of current Test

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Magnetic effects of current Test - 3

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Question 1
1 / -0

A long magnetic needle of length L, having a magnetic moment M and pole strength m units, is broken into two at the middle. The magnetic moment and the pole strength of each piece will respectively be

A
(M/2) and m/2

B
M/2 and m

C
M and m/2

D
M and m

Solution

As the magnetic needle is broken into the two halves from the middle, so the pole strength of the magnetic needle will remain the same. Distance between the two poles becomes halved, so the magnetic moment will become half as magnetic moment is given by the product of the pole strength and the separation between the poles, M = mL
When L' = L/2,
M' = mL/2 = M/2
Question 2
1 / -0

Magnetic field is not associated with

A
a charge in uniform motion

B
an accelerated charge

C
a decelerated charge

D
a stationary charge

Solution

As we know, the magnetic field is produced by the current carrying wire or we can say the source of the magnetic field is current. An accelerated charge, decelerated charge or a charge in uniform motion produce the current whereas the stationary charge produces an electric field.
Question 3
1 / -0

A vertical wire carries a current upwards. The magnetic field at a point due north of the wire is directed

A
upwards

B
due south

C
due west

D
due east

Solution

Using right hand thumb rule, which states that 'If we grasp the conductor in the palm of right hand, so that the thumb points in the direction of the flow of current, the direction in which the fingers curl gives the direction of magnetic field lines', we can conclude that the direction of the magnetic field lines is directed towards west.
Question 4
1 / -0

ABCD is a square loop made of a wire having uniform resistivity. A current enters the loop at A and leaves at D. The magnetic field is

A
zero only at the centre of the loop

B
maximum at the centre of the loop

C
zero at all points outside the loop

D
zero at all points inside the loop

Solution

Here, the current enters at point A, it divides in two parts along AD and along ABCD in the ratio 1: 3 as the current is inversely proportional to the resistance and resistance of length ABCD is 3 times the resistance of AD (R ∝ L) .
The direction of the magnetic field at the centre of the square loop due to the current in AD will be outwards and the direction of the magnetic field due to the current in ABCD will be inwards, which will cancel each other. Hence, the magnetic field at the centre will be zero.
Question 5
1 / -0

AB and CD are long straight conductors, distance d apart, carrying a current I. The magnetic field on BC due to the current in AB and CD

A
is zero at all points

B
is zero only at its midpoint

C
has different magnitudes at different points

D
is maximum at its midpoint

Solution

Using right-hand thumb rule, we can conclude that the direction of the magnetic field at the centre of BC due to AB will be inwards and the direction of the magnetic field at the centre of BC due to CD will also be inwards. Hence, the resultant magnetic field will be added up. So, the value of the magnetic field at the centre of the BC will be maximum.
Question 6
1 / -0

A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated about its axis, then

A
current will be induced in the coil

B
no current will be induced in the coil

C
both emf and current will be induced in the coil

D
None of these

Solution

No current will be induced as there is no change in the magnetic flux linked with coil.
Question 7
1 / -0

A diamagnetic material in a magnetic field moves

A
perpendicular to the field

B
from weaker to the stronger parts of the field

C
from stronger to the weaker parts of the field

D
in none of the above directions

Solution

Option 3 is correct.
Diamagnetic substances are those which are repelled by magnets, and when placed in a non-uniform magnetic field move from stronger to weaker part of the field. Familiar examples of these are bismuth, phosphorus, antimony, copper, water, alcohol and hydrogen.
Question 8
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Materials suitable for making electromagnets should have

A
high retentivity and high coercivity

B
high retentivity and low coercivity

C
low retentivity and low coercivity

D
low retentivity and high coercivity

Solution

The materials suitable for making electromagnets should have high retentivity and low coercivity. Coercivity is usually measured in oersted or ampere/meter units. When the coercive field of a ferromagnet is large, the material is said to be a hard or permanent magnet.
Question 9
1 / -0

A square loop is made by a uniform conductor wire as shown in the figure:

What is the net magnetic field at the centre of the loop, if side length of the square is 'a'?

A

B
Zero

C

D
None of these

Solution

The current will be equally divided at junction P. The field at the centre due to wires PQ and SR will be equal in magnitude but opposite in the direction, so its effective field will be zero. Similarly, the net field due to wires PS and QR is zero. Therefore, the net magnetic field at the centre of the loop is zero.
Question 10
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Two parallel wires carry current in opposite directions. They will

A
attract each other

B
repel each other

C
create gravitational field

D
None of the above

Solution

The two current-carrying conductors attract each other when the currents are in the same direction and repel each other when the currents are in opposite directions.
The two wires experience force in the direction according to Fleming's left hand rule.
Question 11
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An electron moving with a uniform velocity along the positive x-direction enters a magnetic field directed along the positive y-direction. The force on the electron is directed along

A
positive y-direction

B
negative y-direction

C
positive z-direction

D
negative z-direction

Solution

The force on the positive charge moving in the magnetic field is given by Fleming's left hand rule which states that if we stretch the thumb, the centre finger and the middle finger of our left hand such that they are mutually perpendicular to each other, the centre finger gives the direction of current and middle finger points in the direction of magnetic field, then the thumb points towards the direction of the force or motion of the conductor. Direction of the moving electron is opposite to the direction of the thumb. Here, the electron is moving in the positive direction and the magnetic field is directed in the positive y-direction, so the force on the electron will be directed in negative z-direction.
Question 12
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_{The figure shows three situations when an electron with velocity}_{travels through a uniform magnetic field}_{. In each case, what is the direction of magnetic force on the electron?}

A
+ve z-axis, -ve x-axis, +ve y-axis

B
–ve z-axis, -ve x-axis and zero

C
+ve z-axis, +ve y-axis and zero

D
–ve z-axis, +ve x-axis and zero

Solution

According to Fleming's left hand rule, in figures (1) and (2), magnetic force on the electron will be directed in –ve z-axis and –ve x-axis respectively. In figure (3), velocity of electron and direction of magnetic field are antiparallel, so no force will act on the electron.
Question 13
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The north pole of a magnet is brought near a metallic ring as shown in the figure. The direction of induced current in the ring will be

A
clockwise

B
anticlockwise

C
first clockwise and then anticlockwise

D
first anticlockwise and then clockwise

Solution

As the magnet approaches the loop, the current is induced in such a way that it will oppose the motion of the magnet. So, the nearer end of the loop will behave as the north pole and the current in the ring will flow in anticlockwise direction.
Question 14
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The following diagram represents the construction of an ac generator. If the coil rotates in clockwise direction,

A
during the first half rotation, the current in the outer circuit is from B₁ to B₂

B
during the second half rotation, the current in the outer circuit is from B₁ to B₂

C
during both the rotations, the current is from B₁ to B₂

D
None of these

Solution

Using Fleming's right hand rule, which states, 'If the right hand is held with the thumb, first finger and second finger mutually perpendicular to each other (at right angles), the thumb is pointed in the direction of motion of the conductor, and the first finger is pointed in the direction of the magnetic field, then the second finger represents the direction of the induced current.'
Thus, we can say that if the coil rotates in clockwise direction, during the second half rotation, the current in the outer circuit is from B₁ to B₂.
Question 15
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In the arrangement shown in the given figure, there are two coils wound on a non-conducting cylindrical rod. Initially the key is not inserted. Then the key is inserted and later removed. Then,

A
the deflection in the galvanometer remains zero throughout

B
there is a momentary deflection in the galvanometer, but it dies out shortly and there is no effect when the key is removed

C
there are momentary galvanometer deflections that die out shortly; the deflections are in the same direction

D
there are momentary galvanometer deflections that die out shortly; the deflections are in opposite directions

Solution

When the key is closed, there will be a momentary deflection in the coil. When the current becomes constant, the deflection will become zero. And when the key is taken out, current will be induced in opposite direction and the deflection in coil become opposite to the earlier one.
Question 16
1 / -0

An electron and a proton enter a magnetic field with equal velocities. Which one of them experiences more force?

A
Electron

B
Proton

C
Cannot be predicted

D
Both experience the same force

Solution

The force on a moving charge due to the magnetic field is given by F = qvB sinθ.
The magnitude of the charge on the electron and the proton is the same and both are moving with equal velocities. Therefore, magnitude of the force acting on them will be the same.
Question 17
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Magnetism in a magnet is mainly due to

A
orbital motion of the electrons

B
the spin motion

C
the nuclear charge

D
nucleus density

Solution

Magnetism in the magnet is mainly due to the spin motion of the electrons. The magnetic moment consists of two distinct sources. The electron is a particle, which has its own spin which can take on values of ± 1/2 ħ. The nucleus also has protons and neutrons, which have their own intrinsic spins. However, the magnetic moment is inversely proportional to the mass. So, in comparison with the electron, the magnetic moment of the nucleus is minimal.There is also a contribution to the magnetic moment produced by the electrons moving around the nucleus can be considered to act as a current loop. The magnitude of the magnetic moment is related to the angular momentum of the electron travelling around the nucleus.
Question 18
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The two ends of a horizontal conducting rod of length l are joined to a voltmeter. The whole arrangement moves with a horizontal velocity u, the direction of motion being perpendicular to the rod. The vertical component of earth's magnetic field is B. The voltmeter reading is .....

A
Blv only if the rod moves eastward

B
Blv only if the rod moves westward

C
Blv only if the rod moves in any direction

D
zero

Solution

According to Faraday's laws, emf induced across the conducting rod is given by e = Blvsinθ, where θ is the angle between the direction of the magnetic field and the velocity of the rod. Here the direction of the magnetic field is vertically downwards, so if the rod is moving eastward or westward or in any direction in the horizontal plane perpendicular to the direction of the magnetic field, emf is induced.
Question 19
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A geyser of 2.5 KW is connected to a battery of 220 volts. An iron of 1.5 KW is also connected to the battery. To prevent any failure of fuse, the fuse wire must have a current rating of

A
15 A

B
7 A

C
20 A

D
60 A

Solution

Current consumed by the geyser is

Current consumed by the electric iron is

Total current consumed by both is

Hence, fuse rating should be slightly higher than this value (≈ 20 A).
Question 20
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A stream of electrons is projected horizontally to the right. A straight current carrying conductor is supported parallel to the electron stream and above it. If the direction of current in the conductor is from left to right, what will be the effect on the electron stream?

A
The electron beam will be sped up towards the right.

B
The electron stream will be retarded.

C
The electron beam will be pulled downwards.

D
The electron stream will be pulled upwards.

Solution

Magnetic field at a point vertically downwards due to the current carrying wire is directed inwards. As the direction of the beam of current is towards right, so we can say that the current constituted by the beam of electrons is towards left. Using Fleming's left hand rule, we can conclude that the direction of the force on the magnetic field is vertically downwards.