Magnetism and Matter Test 42

Result

Magnetism and Matter Test 42

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Weekly Quiz Competition

Question 1
1 / -0

The potential energy stored in the magnetic field inside the given section of the co-axial cylinder is given by the expression. (by integration of the energy density of the magnetic field over the volume of the section enclosed between the cylindrical shells)?

A
$$\displaystyle U=\frac{\mu_oi^2l}{4\pi}1n\left(\frac{b}{a}\right)$$

B
$$\displaystyle U=\frac{\mu_oi^2l}{2\pi}1n\left(\frac{b}{a}\right)$$

C
$$\displaystyle U=\frac{\mu_oi^2l}{4\pi}\frac{b}{a}$$

D
$$\displaystyle U=\frac{\mu_oi^2l}{2\pi}\frac{b}{a}$$
Question 2
1 / -0

A metal wire of mass $$m$$ slides without friction on two rails at distance $$d$$ apart. The track is in a vertical uniform field of induction $$\vec B.$$ A constant current $$\vec i$$ flows along one rail across the wire and back down the other rail the velocity of the wire as a function of time, assuming it to be at rest initially.

A
$$\dfrac{{Bid}}{m}$$

B
$$\dfrac{{Bidt}}{m}$$

C
$$\dfrac{m}{{Bidt}}$$

D
None of the above

Solution

$$iBl = ma = m\dfrac{{dv}}{{dt}}$$
$$\boxed{\dfrac{{iBl}}{m}dt = v}$$
Question 3
1 / -0

Write the dimension of the magnetic flux in terms of mass, time, length and charge.

A
$$ML^3T^{-1}Q^{-1}$$.

B
$$ML^2T^{-1}Q^{-1}$$.

C
$$ML^4T^{-1}Q^{-1}$$.

D
$$ML^5T^{-1}Q^{-1}$$.

Solution
Question 4
1 / -0

There are two current carrying planar coils each made from identical wires of length L. $$C_1$$ is circular (radius R) and $$C_2$$ is square(side a). They are so constructed that they have same frequency of oscillation when they are place in the same uniform magnetic field $$\vec{B}$$ and carry the same current. Find a in terms of R.

A
$$a=R$$

B
$$a=2R$$

C
$$a=3R$$

D
$$2a=5R$$

Solution

For coil $$C_1$$:
$$n_1=\dfrac{L}{2\pi R}$$, $$m_1=n_1iA_1=\left(\dfrac{L}{2\pi R}\right)$$ (i) $$(\pi R^2)=\dfrac{LiR}{2}$$
$$I_1$$(moment of inertia of the coil $$C_1$$ about an axis through its diameter)
$$=\dfrac{1}{2}MR^2$$
Also, $$\omega_1=\sqrt{\dfrac{m_1B}{I_1}}\left(as T_1=2\pi\sqrt{\dfrac{I_1}{mB}}\right)$$
For the coil $$C_2$$:
As $$n_2=\dfrac{L}{4a}, m_2=n_2iA_2=\left(\dfrac{L}{4a}\right)(i)(a^2)=\dfrac{Lia}{4}$$
$$I_2$$(moment of inertia of coil $$C_2$$ about an axis through its centre and parallel to one of its sides)
$$=\dfrac{1}{12}Ma^2$$
Also, $$\omega_2=\sqrt{\dfrac{m_2B}{I_2}}$$
Since $$\omega_1=\omega_2, \dfrac{m_1}{I_1}=\dfrac{m_2}{I_2}$$
or $$\dfrac{LiR/2}{\dfrac{1}{2}MR^2}=\dfrac{Lia/4}{\dfrac{1}{12}Ma^2}$$ or $$\dfrac{Li}{MR}=\dfrac{3Li}{Ma}$$
Hence, $$a=3R$$.
Question 5
1 / -0

If $$M_2$$ = magnetization of a paramagnetic sample, B = external magnetic field, T = absolute temperature, C = curie constant then according to Curie's law in magnetism, the correct relation is

A
$$M_z = \dfrac{T}{CB}$$

B
$$M_z = \dfrac{CB}{T}$$

C
$$ C = \dfrac{M_z}{T}$$

D
$$ C = \dfrac{T^2}{M_zB}$$

Solution

According to Curie's law:-

$$X\propto\dfrac{1}{T}$$, where $$X$$=magnetic susceptibility and $$T=$$ temperature

$$\implies X=\dfrac{C}{T}$$, where $$C$$=Curie's constant

Magnetization, $$M_z=XB$$

$$\implies M_z=\dfrac{CB}{T}$$

Hence, answer is option-(B).
Question 6
1 / -0

Statement-1: The core of the transformer is laminated to avoid loss of energy.
Statement-2: A laminated metal sheet placed in a changing magnetic field has lower eddy current.

A
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

B
Statement-! is true, statement-2 is false.

C
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

D
Statement-1 is false, statement-2 is true.

Solution

The core of transformer is laminated to avoid loss of energy This statement is true and also a laminated metal sheet placed in a changing $$\vec{B}$$ has lower eddy current.
Question 7
1 / -0

An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in a circular orbit of radius $$R$$.
Obtain an expression for the orbital magnetic moment of the electron.

A
$$\dfrac {e.h}{43\pi m}$$.

B
$$\dfrac {e.h}{5\pi m}$$.

C
$$\dfrac {e.h}{6\pi m}$$.

D
$$\dfrac {e.h}{4\pi m}$$.

Solution
Question 8
1 / -0

As shown in the figure, four identical loops are placed in a uniform magnetic field B. The loops carry equal current i. $$\hat{n}$$ denotes the normal to the plane of each loop. Potential energies in descending order are.

A
I, II, III, IV

B
IV, II, III, I

C
I, III, II, IV

D
IV, III, II, I

Solution
Question 9
1 / -0

A domain in ferromagnetic iron is in the form of a cube of side length $$2$$ $$\mu m$$ then the number of iron atoms in the domain are (Molecular mass of iron $$=55$$g $$mol^{-1}$$ and density $$=7.92$$g $$cm^{-3}$$)

A
$$6.92\times 10^{12}$$ atoms

B
$$6.92\times 10^{11}$$ atoms

C
$$6.92\times 10^{10}$$ atoms

D
$$6.92\times 10^{13}$$ atoms

Solution

The volume of the cubic domain
$$V=(2\mu m)^3=(2\times 10^{-6}m)^3$$
$$=8\times 10^{-18}m^3=8\times 10^{-12}cm^3$$
and mass $$=$$volume $$\times$$ density
$$=8\times 10^{-12}cm^3\times 7.9g cm^{-3}$$
$$=63.2\times 10^{-12}$$g
Now the Avagadro number $$(6.023\times 10^{23})$$ of iron atoms have a mass of $$55$$g. Hence the number of atoms in the domain are.
$$N=\dfrac{63.2\times 10^{-12}\times 6.023\times 10^{23}}{55}=6.92\times 10^{11}$$ atoms.
Question 10
1 / -0

A short bar magnet of magnetic moment m= $$0.32JT^1$$ is placed in a uniform magnetic field of $$0.15 T$$. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

A
(a) $$\overrightarrow {m}$$ parallel to $$\overrightarrow{B}= -mB = -11.8 \times 10^{-2}J:$$ unstable.
(b) $$\overrightarrow {m}$$ anti-parallel to $$\overrightarrow {B}:+mB = +12.8 \times 10^{-2}J$$; stable.

B
(a) $$\overrightarrow {m}$$ parallel to $$\overrightarrow{B}= -mB = -4.8 \times 10^{-2}J:$$ stable.
(b) $$\overrightarrow {m}$$ anti-parallel to $$\overrightarrow {B}:+mB = +4.8 \times 10^{-2}J$$; unstable.

C
(a) $$\overrightarrow {m}$$ parallel to $$\overrightarrow{B}= -mB = -4.8 \times 10^{-2}J:$$ unstable.
(b) $$\overrightarrow {m}$$ anti-parallel to $$\overrightarrow {B}:+mB = +4.8 \times 10^{-2}J$$; stable.

D
(a) $$\overrightarrow {m}$$ parallel to $$\overrightarrow{B}= -mB = -9.8 \times 10^{-2}J:$$ stable.
(b) $$\overrightarrow {m}$$ anti-parallel to $$\overrightarrow {B}:+mB = +10.8 \times 10^{-2}J$$; unstable.

Solution