Magnetism and Matter Test 47

Result

Magnetism and Matter Test 47

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

Three long parallel wires carrying steady currents $$20\ A, 10\ A,10\ A$$ are cut by a perpendicular plane in the vertices $$A$$ and $$B$$ and $$C$$ of a triangle in which angles $$B$$ and $$C$$ are equal. The current of $$20\ A$$ through $$A$$ is in opposite direction through $$B$$ and $$C$$, then:

A
on the line through $$A$$ perpendicular to $$BC$$, the only point at which the magnetic induction vanishes lies on the circumcircle of the triangle$$ABC$$

B
on the line through $$A$$ perpendicular to $$BC$$, the only point at which magnetic induction does not vanish lies on the circumcircle of the triangle $$ABC$$

C
if the triangle $$BAC$$ is equilateral, each side being of length $$10\ cm$$, the magnitude of the mechanical force per unit length on the wire through $$A$$ us zero

D
if the triangle $$ABC$$ is equilateral, each side being of length $$10\ cm$$, the magnitude of the mechanical force per unit length on the wire through $$A$$ is $$1732$$ dyne.

Solution
Question 2
1 / -0

An aircraft is flying at a level height in presence the magnetic field of the Earth. If an electric bulb is connected between the two extreme ends of the wings,

A
a voltage will induce across the wings and the bulb will glow.

B
no voltage will induce across the wings but the bulb will glow.

C
a voltage will induce across the wings but the bulb will not glow.

D
no voltage will induce across the wings and the bulb will not glow.

Solution
Question 3
1 / -0

Consider the situation shown in figure. If the closed loop is completely enclosed in the circuit containing the switch, the closed loop will show____

A
an anticlockwise current-pulse

B
a clockwise current-puls

C
an anti-clockwise current-pulse and then a clockwise current-pulse

D
a clockwise current-pulse and then an anticlockwise current-pulse

Solution
Question 4
1 / -0

A uniform magnetic field along a long cylindrical rod varies with time as $$ B = \alpha t $$ where $$ \alpha $$ is positive constant. the electric field inside the rod as a function of radial distance r from the central axis is proportional to

A
$$ r^2 $$

B
$$ \dfrac {1}{r} $$

C
$$ \dfrac {1}{r^2} $$

D
None of these

Solution
Question 5
1 / -0

An Indian ship with a vertical conducting mass navigates the Indian ocean in the latitude of magnetic equator. To induce the greatest emf in the mast, the ship should proceed:

A
northward

B
southward

C
eastward

D
none of these

Solution
Question 6
1 / -0

A copper rod is bent into a semi-circle of radius a and at ends straight parts are bent along diameter of the semi-circle and are passed through fixed, smooth and conducting ring O and O' as shown in fig. A capacitor having capacitance C connected to the rings. The system is located in uniform magnetic field of induction B such that axis of rotation OO' is perpendicular to the field direction. At initial moment of time t=0 plane of semi circle was normal to the field direction and the semi-circle is set in rotation with constant angular velocity $$\omega$$. Neglect the resistance and inductance of the circuit. The current flowing through circuit as function of time is

A
$$\dfrac {1}{4} \pi \omega^2 a^2 CB cos \omega t$$

B
$$\dfrac {1}{2} \pi \omega^2 a^2CB cos \omega t$$

C
$$\dfrac {1}{4} \pi \omega^2 a^2 CB sin \omega t$$

D
$$\dfrac {1}{2} \pi \omega^2 a^2 CB sin \omega t$$

Solution

When the copper rod is rotated, flux linked with the circuit varies with time.
Therefore, an emf is induced in the circuit.
At time t, the plane of semi-circle makes angle $$\omega$$ t with the plane of the rectangular part of the circuit. Hence, a component of the magnetic induction normal to the plane of the semi-circle is equal to $$B cos \omega t$$

Flux linked with a semicircular part is

$$\phi_1=\dfrac {1}{2} \pi a^2 B cos \omega t$$

Let the area of the rectangular part of the circuit be A.

$$\therefore$$ Flux linked with this part is

$$\phi_2 =BA$$

$$\therefore$$ Total flux linked with the circuit is
$$\phi=\dfrac {1}{2} \pi a^2 B sin (\omega t)$$

Since the resistance of the circuit is negligible, therefore, the potential difference across the capacitor is equal to induced emf in the circuit.

$$\therefore$$ Charge on the capacitor at time t is q=Ce

$$=\dfrac {1}{2} \pi \omega a^2 CB sin (\omega t)$$

But current $$I= \dfrac {dq}{dt}=\dfrac {1}{2} \pi \omega^2 a^2 CB cos (\omega t)$$
Question 7
1 / -0

Two long parallel conducting rails are placed in a uniform magnetic field. On one side, the rails are connected with a resistance R. Two rods MN and M'N' each having resistance r are placed as shown in fig. Now on the rods MN and M'N' forces are applied such that the rods move with constant velocity v.

If in the previous problem resistances of values $$R_1$$ and $$R_2$$ are connected on both ends as shown in Fig. The current I, flowing through resistance $$R_1$$ is given by

A
$$\dfrac {BlR_2(v_1r_2-v_2r_1)}{R_1R_2 (r_1+r_2)+r_2r-1(R_1+R_2)}$$

B
$$\dfrac {BlR_2(v_1r_2+v_2r_1)}{R_1R_2 (r_1+r_2)+r_2r-1(R_1+R_2)}$$

C
$$\dfrac {BlR_2(v_1r_2-v_2r_1)}{R_1R_2 (r_1-r_2)+r_2r-1(R_1-R_2)}$$

D
$$\dfrac {BlR_2(v_1r_2-v_2r_1)}{R_1R_2 (r_1+r_2)+r_2r-1(R_1+R_2)}$$

Solution

From the above figure
$$e_1=Blv_1,, e_2=Blv_2$$

applying kvl to loop,
For (1) $$\rightarrow e_2=(I-I_1+I_2)r_2+I_2R_2$$

For (1) $$\rightarrow e_1+e_2=(I-I_1+I_2)r_2+I_IR_1$$

For (1) $$\rightarrow e_1=Ir_1+I_1R_1$$

the current flowing through $$R_1$$ is

Solve to get $$I_1=\dfrac {BlR_2(v_1r_2-v_2r_1)}{R_1R_2 (r_1+r_2)+r_2r-1(R_1+R_2)}$$
Question 8
1 / -0

A standing wave $$y=2A \sin kx \cos \omega t$$ is set up in the wire AB fixed at both ends by two vertical walls. The region between the walls contains a constant magnetic field B. Now, answer the following question..

In the above equation, the time when the emf becomes maximum for the first time is

A
$$ \dfrac {2\pi} {\omega}$$

B
$$ \dfrac {\pi} {\omega}$$

C
$$ \dfrac {\pi} {2 \omega}$$

D
$$ \dfrac {\pi} {4 \omega}$$

Solution

Given
$$y= 2A sin \,k\,x \,cos \omega t$$

on differentiating we get,

$$v= \dfrac {dy}{dt}=-2A sin kx \omega sin \omega t $$

$$v_{max}=-2A sin kx , \ k= \dfrac {3 \pi}{AB}$$

now,
$$e= \int _0^{t=AB} Bv_{max} dx$$

$$= -2 A \omega B \int _0^{AB} sin kx dx$$

$$= + \dfrac {2\omega AB}{k}[cos \dfrac {3 \pi}{AB} -cos \theta]$$

$$=-\dfrac {4(AB) \omega}{k}$$

$$\omega t = \dfrac {\pi} {2}$$

$$t= \dfrac {\pi} {2 \omega}$$
Question 9
1 / -0

A metal bar is moving with a velocity of $$v=5 cm s^{-1}$$ over a U-shaped conductor.AT $$t=0$$, the external magnetic field is 0.1 T directed out of the page and is increasing at a rate of $$0.2 T s^{-1}$$. Take $$l= 5 cm$$ and $$ t=0, x=5 cm$$

The current flowing in the circuit is

A
2.5 A

B
5 A

C
1 A

D
2 A

Solution

Let us take anticlockwise direction as positive, then the area vector in the upward direction will be positive.
now as we know the flux is
$$\phi= BA=Blx$$

$$e= -{d\phi}{dt}=-l [B \dfrac {dx}{dt}+x \dfrac {dB}{dt}]$$

$$=-5 \times 10^{-2}[ 0.1 \times (-5 \times 10^{-2} + 5 \times 10^{-2} \times 0.2]$$

$$-250 \times 10^{-6} V= -250 \mu V$$

emf is coming out to be negative, so it should be clockwise.

Current $$I= \dfrac {e}{R}= \dfrac {250 \times 10^{-6}}{10^{-4}}=2.5A$$
Question 10
1 / -0

Two long parallel conducting rails are placed in a uniform magnetic field. On one side, the rails are connected with a resistance R. Two rods MN and M'N' each having resistance r are placed as shown in fig. Now on the rods MN and M'N' forces are applied such that the rods move with constant velocity v.

The current flowing through resistance R if the rod MN moves towards left and the rod M'N' moves towards the right is

A
$$\dfrac {Blv}{R+(r/2)}$$

B
$$\dfrac {Blv}{R+r}$$

C
zero

D
$$\dfrac {3Blv}{2(R + r)}$$

Solution

Applying kvl to loop 1
$$-(-I_1-I') R-I_1r +e=0$$ for loop (1)

$$(-I_1+I')=2e$$ for loop (2)

Solve to ger $$I_1=I' \dfrac {e}{R}$$

Hence current in R is zero.
Alternate method,
he EMF induced by moving one slider
$$∣E∣=BvL$$
If both of them in the opposite direction
$$E_{eq}=E-E=0$$
hence,
$$i=0$$