Magnetism and Matter Test 7

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Magnetism and Matter Test 7

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Weekly Quiz Competition

Question 1
1 / -0

A current carrying wire is placed in the grooves of an insulating semi circular disc of radius `R', as shown. The current enters at point A and leaves from point B. Determine the magnetic field at point D.

A

B

C

D
None of these

Solution
Question 2
1 / -0

A pair of stationary and infinitely long bent wires is placed in the X-Y plane as shown in figure. The wires carry currents of 10 A each as shown. The segments L and M are along thex-axis. The segments P and Q are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnetic induction at the origin O.

A
1 × 10^–4 wb/m², towards the reader.

B
1 × 10^–4 wb/m², opposite to the reader.

C
2 × 10^–4 wb/m², towards the reader.

D
Zero

Solution

As point O is along the length of segments L and M, so the field at O due to these segments will be zero.
Also, point O is near one end of a long wire.
The resultant field at O, B_R=B_P+B_Q, this field will be into the plane paper.
⇒B_R= μ₀I/4πRO+ μ₀I/4πSO
But, RO=SO=0.02m
Hence, BR=2× (μ₀/4π) ×(10/0.02)=2×10⁻⁷ (10/0.02) =10⁻⁴Wb m⁻²
Question 3
1 / -0

Determine the magniutde of magnetic field at the centre of the current carrying wire arrangement shown in the figure. The arrangement extends to infinity. (The wires joining the successive square are along the line passing through the centre)

A

B
0

C

D
None of these

Solution
Question 4
1 / -0

Figures shows a long wire bent at the middle to form a right angle. The magnitudes of the magnetic fields at the points P, Q, R and S are B₁, B₂, B₃, B₄respectively. The wire and the circumference of circle are coplaner. T

A
B₁ = B₂ = B₃ = B₄

B
B₁ = 2B₂

C
B₃ = 2B₄

D
none of these
Question 5
1 / -0

Two long parallel wires situated at a distance 2a are carrying equal current `i' in opposite direction as shown in figure. The value of magnetic field at a point P situated at equal distances from both the wires will be :

A

B

C

D

Solution

We know that,
B₁=B₂μ₀/2πr
Angle between B₁ and B₂
BR= √[B₁²+B₂²+2B₁B₂Cos(180-2θ)]
= √2B²+2B²(-Cos2θ)
=B√[2(1-Cos²θ)]
=B√[2(sin²θ)]
=2Bsina
B=(2μ₀i/2πr) x (a/r)
B=(2μ₀ia/2πr²)
Question 6
1 / -0

A uniform beam of positively charged particles is moving with a constant velocity parallel to another beam of negatively charged particles moving with the same velocity in opposite direction separated by a distance d. The variation of magnetic field B along a perpendicular line draw between the two beams is best represented by

A

B

C

D

Solution

Currents are in the same direction, so at the middle B=D
And the B varies exponentially not linearly,
So,
Question 7
1 / -0

Infinite number of straight wires each carrying current I are equally placed as shown in the figure. Adjacent wires have current in opposite direction. Net magnetic field at point P is

A

B

C

D
Zero

Solution

Magnetic field at P is equal to the magnetic field due to AB,CD,EF and so on.
Magnetic field due to AB is +Z direction.
and CD is negative −Z direction and EP is positive +Zdrection and so on.
B_P= (μ₀I/4πd₁) [sin30^°−sin(−30^°)k]+ (μ₀I/4πd₂)[sin30°−sin(−30°)l−k]+ μ0I /4πd₃ [sin30°−sin(−30°)k]+...∞
B_P=2sin30°× (μ₀I/4π) [(1/d₁)−(1/d₂)+(1/d₃)−(1/d₄)+(1/d₅)....∞]k
= μ0I/4π [(1/acos30°)−(1/2acos30°)+(1/3acos30°)−....∞]k
= μ₀I/4π(acos30°) [1−(1/2)+(1/3)−(1/4)+(1/5)....∞]k
=(2μ₀I/4π√3a) ln(1+1)k= (μ₀I/4π√3a) ln2²k=(μ₀I/4π√3a) ln4k
Question 8
1 / -0

Two mutually perpendicular conductors carrying currents I₁ and I₂ lie in one plane. Locus of the point at which the magnetic induction is zero, is a

A
Circle with centre as the point of intersection of the conductor.

B
Parabola with vertex as the point of intersection of the conductors

C
Straight line passing through the point of intersection of the conductors

D
Rectangular hyperbola

Solution

Magnetic field in X-Y plane at position (x,y) due to both wires must be zero
current carrying wire with current i1 is taken along X axis
now the magnetic field at the given point is
B₁= μ₀i₁/2πy
now at the same point the magnetic field due to other wire placed along Y axis is given by,
B₂= μ₀i₂/2πx
now we have to make the magnitude of magnetic field due to both wires same
so, B₁=B₂
μ₀i₁/2πy= μ₀i₂/2πx
now by solving above we have,
y=(i₁/i₂)x
so it’s a straight line with slope i₁/i₂ and passing from origin
Question 9
1 / -0

Current flows through uniform, square frames as shown. In which case is the magnetic field at the centre of the frame not zero ?

A

B

C

D

Solution

According to all figure given,
The magnetic field due to square shape in all options will be zero, Because distribution of current will be less in the ϕ longer path. But its cumulative effect of the magnetic field at the centre will be the same.

In option (A) and (B) current line is passing through the centre, So its perpendicular distance will be zero, so magnetic field in Both (A) and (B) will be zero due to incoming and outgoing current.
Refer image .1

Whereas in case of (D) Magnetic field due to incoming current and outgoing current will be equal and opposite so it will cancel out these effects.
Refer image .2
be in case of (C) Magnetic field will be in same direction with equal magnitude by the incoming and outgoing current, So its effect will not cancel
Option C.
Question 10
1 / -0

The two conductors of a transmission line carry equal current I in opposite directions. The force on each conductor is

A
proportional to 7

B
proportional to X

C
proportional to distance between the conductors

D
inversely proportional to I
Question 11
1 / -0

Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of co-ordinate system. If current I is flowing through each ring then the magnitude of the magnetic field at the common centre is

A

B
Zero

C

D

Solution

So, three rings each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of coordinate system If current I is flowing through each ring them the magnitude of the magnetic field at the common centre is:
Now,
we know,
B for circular loop = ui/2r. This formula is for the magnetic field at the centre of a circular loop of current.
So,
Question 12
1 / -0

Two concentric coils X and Y of radii 16 cm and 10 cm lie in the same vertical plane containing N-S direction. X has 20 turns and carries 16 A. Y has 25 turns & carries 18 A. X has current in anticlockwise direction and Y has current in clockwise direction for an observer, looking at the coils facing the west. The magnitude of net magnetic field at their common centre is

A
5π× 10^–4 T towards west

B
13π × 10^–4 T towards east

C
13π × 10^–4 T towards west

D
5π× 10^–4 T towards east

Solution
Question 13
1 / -0

A current i is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n.

Find the drift velocity v of the electrons.

A

B

C

D

Solution

We know that,
I=V₀nAe
V₀=i/2ae
Question 14
1 / -0

A current i is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n.

If a magnetic field B exists in the region as shown in figure, what is the average magnetic force on the free electrons ?

A

B

C

D

Solution

F=q(VxB)
F=qVBsin θ (Here, θ=90^o,sinθ=1)
F=qVB (Here, V=I/nAe- drift velocity.)
F=ex (I/nAe) x B
F=IB/nA
Question 15
1 / -0

A current i is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n.
Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons.

A

B

C

D
Question 16
1 / -0

A current i is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n.

What will be the potential difference developed across the width of the conductor due to the electron accumulation ? The appearance of a transverse emf, when a current - carrying wire is placed in a magnetic field, is called Hall effect.

A

B

C

D
Question 17
1 / -0

Two circular coils of wire each having a radius of 4 cm and 10 turns have a common axis and are 6 cm apart. If a current of 1 A passes through each coil in the opposite direction find the magnetic induction.
Q.
At the centre of either coil

A
1.3 x 10^-4 T

B
1.3 x 10^-5 T

C
1.2 x 10^-4 T

D
1.2 x 10^-5 T

Solution

I=1A, x=6x10^-2 r=4x10-2m, N=10
B₁= μ₀NIr2/2(r²+x²)^3/2
B₁=2πx10^-7x10x16x10^-4/(16x10^-4+36x10^-4)^3/2
B₁=32x3.14x10^-10/(52)^3/2x10^-6
B₁=2.67x10-5T
Magnetic field due to another coil at centre is,
B₂= μ₀NI/2r=2πx10^-7x10/4x10^-2
Since both are in opposite direction,
B=B₂-B₁
=(1.57-0.267)x10^-4
B=1.3x10^-4T
Question 18
1 / -0

Find the ratio of magnetic field magnitudes at a distance 10 m along the axis and at 60° from the axis, from the centre of a coil of radius 1 cm, carrying a current 1 amp.

A

B

C

D
Question 19
1 / -0

A particle of charge +q and mass m moving under the influence of a uniform electric field E and a magnetic field B enters in I quadrant of a coordinate system at a point (0, a) with initial velocity v and leaves the quadrant at a point (2a, 0) with velocity - 2v . Find
Magnitude of electric field

A

B

C

D
Question 20
1 / -0

A particle of charge +q and mass m moving under the influence of a uniform electric field E and a magnetic field B enters in I quadrant of a coordinate system at a point (0, a) with initial velocity v and leaves the quadrant at a point (2a, 0) with velocity - 2v . Find
Rate of work done by the electric field at point (0, a)

A

B

C

D

Solution

From the work- energy theorem
W=△K.E.
⇒qE.2a=(1/2)m[4υ²−υ²]
⇒E=(3/4)( mυ²/qa)
At P rate of work done by E field will be
=F.υ=(qE)(υ)cos0°
=(3/4)( mυ³/a)