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Magnetic Effects of Current and Magnetism MCQs for NEET — Physics Questions with Answers

Practice free Magnetic Effects of Current and Magnetism (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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Magnetic field lines

Cannot intersect

Are always closed curves

Can pass through vacuum

All of these

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Explanation

Magnetic field lines cannot intersect, are always closed curves, and can pass through a vacuum. Therefore, option o4 which states 'All of these' is the correct answer. Magnetic field lines represent the direction of the magnetic force at a given point and must follow certain rules to accurately depict the magnetic field.

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$If a magnet is suspended at angle 30 ° to the magnetic meridian, the dip needle makes an angle of 45 ° with the horizontal . The value of re a l dip angle at the place is$

$\tan^{- 1} \frac{\sqrt{3}}{2}$

$\tan^{- 1} 1$

$\tan^{- 1} \frac{1}{2}$

$\tan^{- 1} (\frac{2}{\sqrt{3}})$

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Two small bar magnets are placed in air at a distance r apart. The magnetic force between them is proportional to

$r^{2}$

$r^{- 2}$

$r^{- 3}$

$r^{- 4}$

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Explanation

$F = \frac{μ_{0}}{4 π} \frac{6 M_{1} M_{2}}{r^{4}} F α \frac{1}{r^{4}}$

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The vertical component of the earths magnetic field is zero at

Magnetic poles

Magnetic equator

Geographic poles

Geographic equator

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Explanation

$V = B \sin ϕ A t e q u a t o r ϕ = 0 s o V = 0$

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A short magnetic dipole is placed at the origin with its dipole movement directed along the x-axis. If magnetic field induction at a point P (r, 0) is B $\hat{i}$ , the magnetic field induction at point Q(0,2r) will be

$\frac{B}{16} \hat{i}$

$- \frac{B}{8} \hat{j}$

$\frac{B}{16} \hat{j}$

$- \frac{B}{16} \hat{j}$

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Explanation

$At axial point : B = \frac{μ_{0}}{4 π} \frac{2 M}{r^{3}} At equitorial point (at same distance) : B' = \frac{μ_{0}}{4 π} \frac{M}{r^{3}} = \frac{1}{2} \frac{μ_{0}}{4 π} \frac{2 M}{r^{3}} = \frac{1}{2} B At distance 2 r : B' = \frac{μ_{0}}{4 π} \frac{M}{{(2 r)}^{3}} = = \frac{1}{8} \frac{μ_{0}}{4 π} \frac{M}{r^{3}} = \frac{1}{8} \frac{B}{2} = \frac{B}{16}$

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The magnetic field at a point x on the axis of a small bar magnet is equal to the field at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is

$2^{- 3}$

$2^{\frac{- 1}{3}}$

$2^{3}$

$2^{\frac{1}{3}}$

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Explanation

The magnetic field due to a bar magnet follows an inverse cube law, meaning the field strength is inversely proportional to the cube of the distance from the center. If the field is equal at points x and y, their distances from the center must follow the ratio (x/y)^3 = 2. Therefore, the ratio x/y = 2^(1/3).

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A magnet of magnetic moment 20 C.G.S. units are freely suspended in a uniform magnetic field of intensity 0.3 C.G.S. units. The amount of work done in deflecting it by an angle of $30^{o}$ in C.G.S. units is

$3 \sqrt{3}$

$3 (2 - \sqrt{3})$

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Explanation

$W = MB (1 - cosθ) = 20 \times 0.3 \times (1 - \cos 30^{0}) = 6 (1 - \frac{\sqrt{3}}{2}) = 3 (2 - \sqrt{3})$

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Points A and B are situated perpendicular to the axis of a 2cm long bar magnet at large distances X and 3X from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to

1 : 9

2 : 9

27 : 1

9 : 1

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Two short magnets with their axes horizontal to the magnetic meridian are placed with their centres 40 cm east and 50 cm west of the magnetic needle. If the needle remains undeflected, the ratio of their magnetic moments is

4:5

16:25

64:125

2: $\sqrt{5}$

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Explanation

(3)

$M a g n e t i c f i e l d d u e t o a s h o r t b a r m a g n e t o n i t s e q u i t o r i a l l i n e = \frac{μ_{0}}{4 π} (\frac{M}{d^{3}}) A s t h e r e i s n o d e f l e c t i o n i n m a g n e t i c n e e d l e, B_{1} = B_{2} \Rightarrow \frac{μ_{0}}{4 π} (\frac{M_{1}}{{d_{1}}^{3}}) = \frac{μ_{0}}{4 π} (\frac{M_{2}}{{d_{2}}^{3}})$

$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3} = {(\frac{40}{50})}^{3} = \frac{64}{125}$

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If a bar magnet of magnetic moment M is freely suspended in a uniform magnetic field of strength B, the work done in rotating the magnet through an angle is

$M B (1 - \sin θ)$

$M B \sin θ$

$M B \cos θ$

$M B (1 - \cos θ)$

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Explanation

(d)

$W = M B (\cos θ_{1} - \cos θ) = M B (\cos 0 - \cos θ) = M B (1 - \cos θ)$

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