Electromagnetic Induction Test

Result

Electromagnetic Induction Test - 9

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

In an iron cored coil the iron core is removed so that the coil becomes an air cored coil. The inductance of the coil will

A
increase

B
decrease

C
remain the same

D
initially increase and then decrease

Solution

In an iron cored coil the iron core is removed so that the coil becomes an air cored coil. The inductance of the coil will decrease.
Question 2
1 / -0.25

The magnetic flux through a stationary loop with resistance R varies during interval of time T as f = at (T ^_ t). The heat generated during this time neglecting the inductance of loop will be

A

B

C

D

Solution

E=dx/dt =d(a(Tt-t² ))/dt=a(T-2t)
I=E/R=a(T-2t)/R
Heat liberated,
△k=I² Rdt

=(a² /R)[T² t+(4t³ /3)-2Tt² ]^T₀
=△H=a² T³ /3R
Question 3
1 / -0.25

The dimensions of permeability of free space can be given by

A
[MLT^-2 A^-2 ]

B
[MLA^-2 ]

C
[M^L-3 T² A² ]

D
[MLA^-1 ]

Solution

In SI units, permeability is measured in Henries per meter H/m or Hm −1.
Henry has the dimensions of [ML² T⁻² A⁻² ].
Dimensions for magnetic permeability will be [ML² T⁻² A⁻² ]/[L]=[MLT⁻² A⁻² ]
Question 4
1 / -0.25

A closed planar wire loop of area A and arbitrary shape is placed in a uniform magnetic field of magnitude B, with its plane perpendicular to magnetic field. The resistance of the wire loop is R. The loop is now turned upside down by 180 °so that its plane again becomes perpendicular to the magnetic field. The total charge that must have flowen through the wire ring in the process is

A
< AB/R

B
= AB/R

C
= 2AB/R

D
None

Solution

Induced emf= d(flux)/dt =R[dq /dt]
⇒Δq= Δ(flux)/R=2BA/R
Question 5
1 / -0.25

A square coil ABCD is placed in x-y plane with its centre at origin. A long straight wire, passing through origin, carries a current in negative z-direction. Current in this wire increases with time.The induced current in the coil is

A
clockwise

B
anticlockwise

C
zero

D
alternating

Solution

As magnetic field lines due to current carrying wire is along the surface and hence,
ϕ=B.A=BAcos90^O =0
Question 6
1 / -0.25

Two identical coaxial circular loops carry a current i each circulating in the same direction. If the loops approach each other

A
the current in each will decrease

B
the current in each will increase

C
the current in each will remain the same

D
the current in one will increase and in other will decrease

Solution

As the two identical coaxial circular loops carry a current i each circulating in the same direction, their flux add each other. When the two are brought closer, the total flux linkage increases. By Lenz 's law a current will be induced in both the loops such that it opposes this change in flux. As the flux increases, the current in each loop will decrease so as to decrease the increasing flux.
Question 7
1 / -0.25

In the arrangement shown in given figure current from A to B is increasing in magnitude. Induced current in the loop will

A
have clockwise direction

B
have anticlockwise direction

C
be zero

D
oscillate between clockwise and anticlockwise

Solution

The direction of the induced current is as shown in the figure, according to Lenz ’s law which states that the indeed current flows always in such a direction as to oppose the change which is giving rise to it.
Question 8
1 / -0.25

The north pole of a magnet is brought near a coil. The induced current in the coil as seen by an observer on the side of magnet will be

A
in the clockwise direction

B
in the anticlockwise direction

C
initially in the clockwise and then anticlockwise direction

D
initially in the anticlockwise and then clockwise direction

Solution

The direction of the current will be anticlockwise.

According to Lenz 's law, the current in the coil will be induced in the direction that 'll oppose the external magnetic field .
As the flux due to the external magnet is increasing ( as the N-pole is brought close ) , the coil will have to induce current in the anticlockwise direction to oppose this increase in flux and not in clockwise direction as that 'll end up supporting the external flux.
Question 9
1 / -0.25

A metal sheet is placed in a variable magnetic field which is increasing from zero to maximum. Induced current flows in the directions as shown in figure. The direction of magnetic field will be -

A
normal to the paper, outwards.

B
normal to the paper, inwards

C
from east to west

D
from north to south

Solution

The direction of these circular magnetic lines is dependent upon the direction of current. The density of the induced magnetic field is directly proportional to the magnitude of the current. Direction of the circular magnetic field lines can be given by Maxwell 's right hand grip rule or Right handed cork screw rule.
Question 10
1 / -0.25

A small conducting rod of length l, moves with a uniform velocity v in a uniform magnetic field B as shown in fig-

A
Then the end X of the rod becomes positively charged

B
the end Y of the rod becomes positively charged

C
the entire rod is unevely charged

D
the rod becomes hot due to joule heating

Solution

The rod is moving towards the right in a field directed into the page.
Now, if we apply Fleming 's right hand rule, then the direction of induced current will be from end X to end Y.
But, according to Lenz 's law the emf induced in the rod will be such that it opposes the motion of the rod.
Hence, the actual emf induced will be from end Y to end X. So, the current will also flow from end Y to end X.
Now, using the convention of current end Y should be positive and end X should be negative.
So, correct answer is option b