Magnetism and Matter Test 31

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Magnetism and Matter Test 31

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Question 1
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A long magnet moves with constant velocity along the axis of a fixed metal ring. It starts from a large distance from the ring, passes through the ring and then moves away far from the ring. The current in the ring is plotted against time. Which of the following best represents the resulting curve?

A

B

C

D

Solution

As the bar magnet passes through the coil, the flux linked with the ring first increases and then decreases. The current in the ring will therefore reverse in direction as the magnet passes through it. Also, the rate of change of flux is maximum when the magnet is close to the coil therefore the magnitude of current when the magnet is near coil is maximum.
Question 2
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If the angular momentum of an electron of mass $$m$$ is $$J$$ then the magnitude of the magnetic moment will be

A
$$\dfrac{eJ}{m}$$

B
$$\dfrac{eJ}{2m}$$

C
$$2eJm$$

D
$$\dfrac{2m}{eJ}$$

Solution

Magnetic moment, $$M=I\times A...........(1)$$ where I is current and A is area

$$ I=\dfrac{e}{T}=\dfrac{e}{\dfrac{2\pi r}{v}}=\dfrac{ev}{2\pi r} $$

$$ A=\pi {{r}^{2}} $$

Angular Momentum $$\,J=mvr.......(2) $$

Put values of $$I$$ and $$A$$ in equation (1)

$$ M=\dfrac{ev}{2\pi r}\pi {{r}^{2}} $$

$$ M=\dfrac{evr}{2} $$

$$ M=\dfrac{eJ}{2m}\,\,\,\,\,\,\,\,\,\left( \because vr=\dfrac{J}{m} \right) $$
Question 3
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A magnetising force of $$360 Am^{-1}$$ produces a magnetic flux density of $$0.6 T$$ in a ferromagnetic material. The susceptibility of the material is:

A
$$1625$$

B
$$1329$$

C
$$2105$$

D
$$1914$$

Solution

Permeability of material, $$\mu = \frac{B}{\mu } = \frac{{0.6}}{{360}} = 1.67 \times {10^{ - 3}}{T^{ - 1}}\,Am$$
$$\mu = {\mu _0}\left( {1 + {X_m}} \right)$$
$$\therefore $$ Susceptibility, $${X_m} = \frac{\mu }{{{\mu _0}}} - 1 = \frac{{1.67 \times {{10}^{ - 3}}}}{{47 \times {{10}^{ - 7}}}} - 1 = 1329$$
Hence, Option $$B$$ is correct.
Question 4
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An electron beam is moving near to a conducting loop then the induced current in the loop:-

A
clockwise

B
anticlockwise

C
both

D
no current

Solution

The induced current in a loop moves from right to left. Since the electrons flow in direction opposite to that of the current thus the electron beam will move from left to right (anticlockwise).
Question 5
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If two straight current carrying wires are kept perpendicular to each other almost touching, then the wires

A
Attract each other

B
Repel each other

C
Remain Stationary

D
Become parallel to each other

Solution

Solution: Let given figure shows two long current carrying conductor placed $$\perp^{rd}$$ to each other.
Here $$F_1$$ & $$F_2$$ be equal in magnitude and opposite in direction. It will provide torque & conductor will rotate & become parallel.
$$\Rightarrow$$ Even after that it may be attracted or repelled depend upon-direction of current, but initially it become parallel to $$(1)$$.
Question 6
1 / -0

Copper is :

A
Paramagnetic

B
Diamagnetic

C
Ferromagnetic

D
None of these

Solution
Question 7
1 / -0

In a long hollow vertical metal pipe a magnet is dropped. During its fall, the acceleration of magnet

A
will decrease linearly

B
will decrease upto a value which is less than $$g$$.

C
will decrease to zero and will attain a terminal speed

D
may increase or decrease

Solution

The falling magnet will experience an upward force and
so the acceleration of magnet will decreases and then the magnet will a quire a terminal speed.
Hence $$C$$ is the correct answer.
Question 8
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The wing span of an aeroplane is 20 meter. it is flying in a field.where the vertical component of magnetic field of earth is $$ 5 \times 10^-5 $$ tesla with velocity $$ 360mh^1 $$. The potential difference product between the blades is:

A
0.15 V

B
0.16 V

C
0.20 V

D
0.30 V

Solution

Given,

Wing span, $$L=20\,m$$

Magnetic Field $$\overset{\scriptscriptstyle\rightharpoonup}{B}=\ 5\times {{10}^{-5}}\ T$$

$$1\ mile=1.6\ kilometer$$

Velocity, $$v=360\ miles/h=\ 360\times 1.6\ =576\ km/s$$

$$v=\ 160\ m/s$$

Potential Difference, $$V=BLv=5\times {{10}^{-5}}\times 20\times 160=0.16\ V$$
Potential difference is $$0.16\,V$$
Question 9
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A bar magnet of moment $$2$$ $$A-m^2$$ is free to rotate about a Vertical axis passing through its centre. The magnet is released form rest from east west direction. The $$K.E.$$ of the magnet as it takes north-south direction is ($$B_H = $$25$$X 10^{-6}T$$)

A
$$25$$ x $$10^{-6}$$ J

B
$$50$$ x $$10^{-6}$$ J

C
$$75$$ x $$10^{-6}$$ J

D
$$100$$ x $$10^{-6}$$ J

Solution

Given:
Magnetic moment, m= 2Am$$^{2} $$
B$$_H = 25 ×10^{-6} T $$
$$\theta_1 = \dfrac{\Pi}{2}$$
$$\tgeta_2 = 0
Solution:
$$U_1= -mBCos\theta_1$$
$$\Rightarrow U_1= - 2× 25×10^{-6}Cos\dfrac{\Pi}{2}$$
$$\Rightarrow U_1 =0$$
$$U_2 =-mBCos\theta_2$$
$$\Rightarrow U_2 =-2×25×10^{-6}×Cos0$$
$$\Rightarrow U_2 = -50×10^{-6} $$
Change in K.E. = $$U_1-U_2 = 0-(-50×10^{-6})$$
$$\Rightarrow Change in K.E. = 50×10^{-6} J $$
Hence, option B is correct.
Question 10
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A current is flowing through a thin cylindrical shell of radius R. If the energy density in the medium, due to the magnetic field, at a distance r from the axis of the shell is equal to U then which of the following graphs is correct?

A

B

C

D

Solution

When a current flows through cylindrical shell, then according to Ampere circuital law, magnetic induction inside it will be equal to zero. Hence energy density at $$r<R$$ is eual to zero.

Therefore, $$(a),(c)$$ and $$(d)$$ are wrong.

When $$r>R, B=\dfrac{\mu_{0}i}{2\pi r}$$

since $$U=\dfrac{B^{2}}{2\mu_{0}},$$ therefore, outside the shell, $$U=\dfrac{\mu_{0}i^{2}}{9\pi^{2}r^{2}}.$$

It means, just outside the shell, $$U=\dfrac{\mu_{0}i^{2}}{9\pi^{2}R^{2}}$$ and when $$r\rightarrow \infty, U\rightarrow 0.$$
Hence $$(b)$$ is correct.