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Physics 2 Marks Questions

Magnetism and Matter · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 11 questions.

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Question 12 Marks
What is paramagnetic substance?
Answer
The substance which on keeping in any magnetic field are slightly magnetized in the direction of the field and are attracted towards the ends of a powerful magnet when brought near it are known as paramagnetic substances and this property is known as Para magnetism.
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Question 22 Marks
How magnetic field lines are drawn? Write their four characteristics.
Answer
Magnetic field lines are drawn with the help of iron fillings or magnetic needle. Four main characteristics are :
(1) More the number of force lines passing through a point, more strong will be magnetic field at that place.
(2) Two force lines never intersect each other.
(3) Tangent drawn at any point on the field lines gives the direction of magnetic field at that point.
(4) They are directed from north pole to south pole outside the magnet and inside the magnet, their direction is from south pole to north pole.
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Question 32 Marks
What are diamagnetic substances? Give one example.
Answer
Those substances which on keeping in any magnetic field are attracted a bit towards the opposite side of the magnetic field and when brought near the ends of a powerful magnet are somewhat repelled, are known as diamagnetic substances. This property of these substances is known as diamagnetism. Examples are zinc (Zn), copper (Cu), silver (Ag), gold (Au) etc.
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Question 42 Marks
Name two substances out of which magnetic susceptibility of one is positive and of other negative. What does negative magnetic susceptibility represent?
Answer
Magnetic susceptibility of paramagnetic substances (such as aluminium, calcium etc.) is positive and of diamagnetic substances (such as copper, lead etc.) is negative. Negative magnetic susceptibility exhibits that these substances are magnetised in opposite direction of external magnetic field.
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Question 52 Marks
A bar magnet is kept in a uniform magnetic field such that its magnetic moment is at an angle with the direction of $\overrightarrow{ B }$. Write an expression for potential energy of this system. When it will be minimum?
Answer
Direction of magnetic moment of a bar magnet is along its axis and it is making an angle with $\overrightarrow{ B }$.
(\therefore$ Potential energy in this condition, $U =-m B \cos \theta$ For minimum of U ,
$ \theta=0^{\circ} \text { and } U_{\min }=-m B \cos 0^{\circ}=-m B $
Hence, when $\vec{M}$ and $\vec{B}$ are parallel, then $U$ will be minimum.
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Question 62 Marks
Write magnitude and direction of force acting on poles of bar magnet placed in a magnetic field.
Answer
When a bar magnet is suspended independently in any magnetic field, then in stable condition, a force acts on its both poles. Its direction is along the direction of magnetic field. On north pole its direction is towards the magnetic field and at south pole, it is in opposite direction to that of the magnetic field. Magnitude of this force is $\overrightarrow{ F }=q_{ m } \overrightarrow{ B }$. Hence, unit of $q_m$ will be newton/tesla.
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Question 72 Marks
If the bar magnet in Q. 58 is turned around by 180°, where will the new null points be located?
Answer
When the bar magnet is turned through 180°, neutral points would lie on equatorial line, so that
$B_2=\frac{\mu_0}{4 \pi} \frac{M}{d_2^3}=H$...(1)
From question 5.8,
$B_1=\frac{\mu_0}{4 \pi} \frac{2 M}{d_1^3}=H$
From eqn. (1) and (2),
$\frac{\mu_0}{4 \pi} \frac{2 M}{d_2^3}=\frac{\mu_0}{4 \pi} \frac{2 M}{d_1^3}$
$\therefore \quad d_2^3=\frac{d_1^3}{2}$
$=\frac{(14)^3}{2}$
$d_2=\frac{14}{2^{1 / 3}}=11.1 cm$

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Question 82 Marks
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-sourth direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null points (i.e. 14 cm) from the centre of the magnet? (at null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field).
Answer
As null points are on the axis of the magnet, therefore,
$B_1=\frac{\mu_0}{4 \pi} \frac{2 M}{d^3}=H$
On the equatorial line of magnet at same direction (d), field due to the magnet is
$B_2=\frac{\mu_0}{4 \pi} \frac{M}{d^3}=\frac{B_1}{2}=\frac{H}{2}$
$\therefore$ Total magnetic field at this point on equatorial line
$B = H + B _2= H +\frac{ H }{2}$
$=\frac{3}{2} H =\frac{3}{2} \times 0.36=0.54 G$
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Question 92 Marks
If the solenoid in question 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, then what is the magnitude of torque on the solenoid when its axis makes an angle of$30^{\circ}$ with the direction of applied field?
Answer
Given Uniform magnetic field B = 0.25 T ,$\theta=$ $30^{\circ}$, Magnitude of torque on the solenoid m = 0.60 Am² (Taken from Q. 5.3.)
$\tau=$ Magnitude of twisting torque on the solenoid, whose value we have to find
$\tau=m B \sin \theta$
$=0.60 \times 0.25 \times \sin 30^{\circ}$
$=0.60 \times 0.25 \times \frac{1}{2}$
$\begin{array}{l}=0.075 Nm \\
=7.5 \times 10^{-2} J\end{array}$
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Question 102 Marks
A closely wound solenoid of 800 turns and area of cross-section $2.5 \times 10^{-4} m^2$ carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer
Given Number of turns N = 800, Area of cross-section (A) = $( A )=2.5 \times 10^{-4} m^2$,Current in the solenoid l = 3.0 A Associated magnetic moment m = ?
For current carried solenoid m = NIA
On putting values,
$m =800 \times 3 \times 2.5 \times 10^{-4}$
$=60 \times 10^{-2} Am ^2$
$=0.60 Am ^2=0.60 JT ^{-1}$
It is along the direction of the axis of the solenoid. In this way, a current carrying solenoid behaves like a bar magnet.
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Question 112 Marks
A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5 \times 10^{-2} J$. What is the magnitude of magnetic moment of the magnet?
Answer
Given: Uniform external magnetic field B = 0.25 T, $\theta=30^{\circ}$,Torque of magnitude t =$=4.5 \times 10^{-2} J$
Magnitude of magnetic moment of magnet m = ?
We know that torque of magnitude $\tau= mB \sin \theta$
$\therefore$ Magnetic moment $m=\frac{\tau}{ B \sin \theta}$
On putting values, $m=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$
$m=\frac{4.5 \times 10^{-2}}{0.25 \times \frac{1}{2}}$
$m=\frac{9}{25}=0.36 JT ^{-1}$

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