Magnetism and Matter Test 35

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Magnetism and Matter Test 35

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Question 1
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A magnet of magnetic moment $$\bar M$$ moves with velocity $$\vec v$$ towards a magnet. Consider a small circular loop whose plane is normal to $$\vec v$$. Its radius $$r$$ is so small that magnetic induction is almost constant over it. Then :

A
the magnetic flux through the area of the loop is constant

B
the electric field intensity along the tangent to the loop and in the plane of the loop is of magnitude $$3\mu_0\dfrac{Mvr}{4\pi x^4}$$ and direction $$\vec E\bot \vec v$$

C
the electric field intensity is in the direction along $$\vec v$$

D
the electric field intensity is not induced anywhere

Solution

As the magnetic flux is given by:
$$ \Phi = \overrightarrow{B} \cdot \overrightarrow{A} $$
And since magnetic induction is almost constant over the loop, the magnetic force of Bar Magnet with magnetic moment $$ \overrightarrow{M} $$, at a distance $$ x $$ is given by:
$$ \overrightarrow{B} = \dfrac{\mu_0}{4 \pi} \dfrac{2 \overrightarrow{M}}{x^3} $$
Thus with a loop of area, $$ A = \pi r^2 $$,
$$ \Phi = \dfrac{\mu_0}{4 \pi} \dfrac{2 M}{x^3} \pi r^2 $$

Now induced EMF, $$ \varepsilon $$,
$$ \varepsilon = \int{\overrightarrow{E} \cdot \overrightarrow{dl}} = -\dfrac{\partial \Phi}{\partial t} $$
Thus,
$$ \varepsilon = E \cdot (2 \pi r) = -\left(-3 \dfrac{\mu_0}{4 \pi} \dfrac{2M}{x^4} \dfrac{dx}{dt} \pi r^2 \right) $$
But $$ v = -\dfrac{dx}{dt} $$,

Therefore,
$$ E = -3 \mu_0 \dfrac{Mvr}{4 \pi x^4} $$
The negative sign just shows that the induced Electric field opposes the change in Magnetic field, since the electric field is along the circumference of the loop, it is perpendicular to the velocity of the magnet.

Thus, option (B) is correct.
Question 2
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Which of the following is correct representation of magnetic lines of force ?

A

B

C

D
None of the above

Solution

The lines of magnetic field from a bar magnet form closed lines. By convention, the field direction is taken to be outward from the north pole and in to the south pole of the magnet. Inside a bar magnet its from south to north pole.

Hence option A is correct.
Question 3
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an equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the paper. At time t=0, side DC of the loop is at edge of the magnetic field.The induced current I(i) versus time (t) graph will be

A

B

C

D

Solution

Let 2a be the side of the triangle and b be the length AE.

$$\dfrac {AH}{AE}=\dfrac {GH}{EC}\Rightarrow GH=(\dfrac {AH}{AE})EC$$

or $$GH=\dfrac {(b-vt)}{b}a=a-(\dfrac {a}{b}vt)$$

$$\therefore FG=2GH =2[a-\dfrac {a}{b}vt]$$

$$\therefore$$ Induced emf, , $$e=BV(FG)=2Bv(a-\dfrac {a}{b}vt)$$

$$\therefore$$ induced current, $$i=\dfrac {e}{R}=\dfrac {2Bv}{R}[a-\dfrac {a}{b}vt]$$

or $$i=K_1-K_2 t$$
Thus i-t graph is a straight line with a negative slope and positive intercept.
Question 4
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A square conducting loop, 20.0 cm on a side is placed in the same magnetic field as shown in Fig. Centre of the magnetic field region, where $$db/dt =0.035 T s^{-1}$$

The current induced in the loop if its resistance is $$2.00 \Omega$$ is

A
$$2.50 \times 10^{-4}A$$

B
$$4.35 \times 10^{-4}A$$

C
$$1.25 \times 10^{-4}A$$

D
$$7.37 \times 10^{-4}A$$

Solution

As we know that
$$I=\dfrac {\epsilon}{R}=\dfrac {A}{R} $$

$$\dfrac {dB}{dt}=\dfrac {L^2}{R}dfrac {dB}{dt}$$

$$= \dfrac {(0.20 m)^2 (0.0350 T/s)}{1.90 \omega}$$

$$7.37 \times 10^{-4}A$$
Question 5
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A rod lies across frictionless rails in a uniform magnetic field $$\overrightarrow {B}$$ as shown in fig. The rod moves to the right with speed V. In order to make the induced emf in the circuit to be zero, the magnitude of the magnetic field should

A
not change

B
increase linearly with time

C
decrease linearly with time

D
decrease nonlinearly with time

Solution

The magnetic flux link is constant with the time.
$$Blvt =constant$$
Hence: $$B=\dfrac {C}{lVt}$$
Option D.
Question 6
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A square coil ACDE with its plane vertical is released from rest in a horizontal uniform magnetic field $$\overrightarrow { B } $$ of length 2 L The acceleration of the coil is

A
less than g for all the time till the loop crosses the magnetic field completely

B
less than g when it enters the field and greater than g when it comes out of the field.

C
g all the time

D
less than g when it enters and comes out of the field but equal to g when it is within the field

Solution

When the coil is within the field, there is no; change in the magnetic flux passing through it. Thus, no current will be induced and the acceleration will be g. But according to Lenz's law, the induced current will oppose its motion when it enters or leaves the field. Therefore, acceleration will be less than g.
Question 7
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A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field is increasing at a constant rate of $$dB/dt Ts^{-1}$$. An electron placed at the point P on the periphery of the field, experiences an acceleration

A
$$\dfrac {1}{2} \dfrac {eR}{m}{dB}{dt}$$ towards left

B
$$\dfrac {1}{2} \dfrac {eR}{m}{dB}{dt}$$ towards right

C
$$ \dfrac {eR}{m}{dB}{dt}$$ towards left

D
zero

Solution

At P induced electric field,

$$\int E.dl=-A\dfrac{dB}{dt}$$

$$E(2\pi R)=-(\pi R^2) \dfrac{dB}{dt}$$

The induced electric field at point P:

$$E= \dfrac {R}{2}\dfrac {dB}{dt}$$ towards right

acceleration of electron: $$a= \dfrac {eE}{m}=\ \dfrac {eR}{2m}\dfrac {dB}{dt}$$ towards left.
Question 8
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The magnetic induction at the center $$O$$ in the figure shown is

A
$$\dfrac {\mu_0i}{4}\left( \dfrac {1}{R_1}-\dfrac {1}{R_2}\right)$$

B
$$\dfrac {\mu_0i}{4}\left( \dfrac {1}{R_1}+\dfrac {1}{R_2}\right)$$

C
$$\dfrac {\mu_0i}{4}\left( R_1-R_2\right)$$

D
$$\dfrac {\mu_0i}{4}\left( R_1+R_2\right)$$

Solution

In the following figure, magnetic fields at $$O$$ due to sections $$1, 2, 3$$ and $$4$$ are considered as $$B_1, B_2, B_3$$ and $$B_4$$ respectively.
$$B_1=B_3=0$$

$$B_2=\dfrac{\mu_0}{4\pi}.\dfrac{\pi i}{R_1}\otimes $$

$$B_4=\dfrac{\mu_0}{4\pi}.\dfrac{\pi i}{R_2}\odot$$

So $$B_{net}=B_2-B_4\Rightarrow B_{net}=\dfrac{\mu_0i}{4}\left( \dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \otimes $$
Question 9
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In the figure, shown the magnetic induction at the center of their arc due to the current in the potion $$AB$$ the magnetic induction at $$O$$ due to the whole length of the conductor is

A
$$\dfrac {\mu_0i}{r}$$

B
$$\dfrac {\mu_0i}{2r}$$

C
$$\dfrac {\mu_0i}{4r}$$

D
$$Zero$$

Solution

The induction due to $$AB$$ and $$CD$$ will be zero. Hence the whole induction will be due to the semicircular part $$BC$$.
$$B=\dfrac{\mu_0i}{4r}$$
Question 10
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In the figure shown, the magnetic induction at the center of their arc due to the current in the potion $$AB$$ will be

A
$$\dfrac {\mu_0i}{r}$$

B
$$\dfrac {\mu_0i}{2r}$$

C
$$\dfrac {\mu_0i}{4r}$$

D
$$Zero$$

Solution

The magnetic induction at $$O$$ due to the current in portion $$AB$$ will be zero because $$O$$ lies on $$AB$$ when extended.