NEET Magnetic Properties of Matter MCQs & PYQs

Practice NEET Magnetic Properties of Matter Questions

Question 31:

easy

Consider the following statements:

Statement (A): All magnetism is due to circulating currents i.e. there cannot be magnetic field without current.

Statement (B): The orbital magnetic moment of an electron in the ground state of a hydrogen atom is equal to one Bohr Magneton.

Choose the correct option.

Statement (A) is incorrect because elementary particles like electrons have intrinsic spin magnetic moments not due to actual circulating currents. Statement (B) is correct as $\mu_l = 1 \mu_B$ for ground state.

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Question 32:

moderate

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their center is the same?

(2006)

Magnetic field $B = \frac{\mu_0 N I}{2 R}$. Since $B$ and length of wire are same, $N_1 R_1 = N_2 R_2$. With $R_1 = 2R_2$, $N_1 = N_2 / 2$. The potential difference ratio $\frac{V_1}{V_2} = \frac{I_1 R_{\text{wire}, 1}}{I_2 R_{\text{wire}, 2}} = 4$.

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Question 33:

moderate

A coil in the shape of an equilateral triangle of side $L$ is suspended between the pole pieces of a permanent magnet such that $B$ is in plane of the coil. If due to a current $i$ in the triangle a torque $tau$ acts on it, the side $L$ of the triangle is :

(2005)

The magnetic moment is $M = i A = i \left( \frac{\sqrt{3}}{4} L^2 \right)$. The torque is $\tau = M B = i \frac{sqrt{3}}{4} L^2 B$. Solving for $L$ yields $2 \left[ \frac{\tau}{\sqrt{3}Bi} \right]^{1/2}$.

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Question 34:

easy

A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes :

(1988)

In stable equilibrium, the magnetic moment vector aligns parallel to the magnetic field direction. Consequently, the plane of the coil becomes perpendicular to the magnetic field.

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Question 35:

easy

A uniform conducting wire of length $12\text{ a}$ and resistance $R$ is wound up as a current carrying coil in the shape of, i. an equilateral triangle of side $a$. ii. a square of side $a$. The magnetic dipole moments of the coil in each case respectively are:

(2021)

For triangle, $N_1 = 4$, $A_1 = \frac{\sqrt{3}}{4}a^2$, so $M_1 = \sqrt{3}Ia^2$. For square, $N_2 = 3$, $A_2 = a^2$, so $M_2 = 3Ia^2$.

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Question 36:

easy

A wire of length $L\text{ m}$ carrying a current of $I\text{ A}$ is bent in the form of a circle. Its magnetic moment is:

(2020-Covid)

Radius $r = L / (2\pi)$. Area $A = \pi r^2 = L^2 / (4\pi)$. Magnetic moment $M = I A = I L^2 / (4\pi)$.

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Question 37:

easy

A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by:

(2007)

Equivalent current $I = q / T = qv / (2\pi R)$. Magnetic moment $\mu = I A = \frac{qv}{2\pi R} \pi R^2 = \frac{qvR}{2}$.

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Question 38:

easy

If number of turn, area and current through it is given by $n$, $A$ and $i$ respectively then its magnetic moment will be:

(2001)

Magnetic dipole moment of a current-carrying coil is the product of number of turns, current, and area, i.e., $M = niA$.

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Question 39:

easy

4. A bar magnet of magnetic moment $M$ is cut into two parts of equal length. The magnetic moment of each part will be (1997)

When a bar magnet is cut into two equal parts perpendicular to its length, the length of each piece becomes $L/2$ while the pole strength $m$ remains unchanged.
New magnetic moment $M' = m \times (L/2) = M/2 = 0.5M$.

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Question 40:

easy

5. A closely wound solenoid of 2000 turns and area of cross section $1.5 \times 10^{-4}\text{ m}^2$ carries a current of $2.0\text{ A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2}\text{ tesla}$ making an angle of $30^\circ$ with the axis of the solenoid. The torque on the solenoid will be (2010 Mains)

Magnetic dipole moment $M = NIA = 2000 \times 2.0 \times 1.5 \times 10^{-4} = 0.6\text{ A m}^2$.
Torque $\tau = MB \sin\theta = 0.6 \times (5 \times 10^{-2}) \times \sin 30^\circ = 1.5 \times 10^{-2}\text{ Nm}$.

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