Magnetic Effect Of Current JEE Advanced Previous Year Questions With Solutions

When a current flows through a conductor, a magnetic field is produced in the region surrounding the conductor. This effect is known as the magnetic effect of electric current. In the year 1820, Prof. H.C. Oersted observed that a tiny magnetic needle placed above or below a straight conductor deflected when a current was passed through the conductor. The direction of the deflection is given by Ampere’s swimming law.

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Question 1) A hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle?

\(\begin{array}{l}(A)\ \frac{\mu _{0}I}{4\pi r}(2-\pi )\end{array} \)

\(\begin{array}{l}(B)\ \frac{\mu _{0}I}{4\pi r}(2+\pi )\end{array} \)

\(\begin{array}{l}(C)\ \frac{\mu _{0}I}{2\pi r}(2+\pi )\end{array} \)

\(\begin{array}{l}(D)\ \frac{\mu _{0}I}{2\pi r}(2-\pi )\end{array} \)

Answer (B)

Solution

Magnetic field due to each of the straight wire

\(\begin{array}{l}=\frac{\mu _{0}I}{4\pi r}\end{array} \)

Magnetic field due to semicircular arc

\(\begin{array}{l}=\frac{\mu _{0}I}{4r}\end{array} \)

Thus, the total magnetic field at point P

\(\begin{array}{l}B=\frac{\mu _{0}I}{4\pi r}+\frac{\mu _{0}I}{4\pi r}+\frac{\mu _{0}I}{4r}\end{array} \)

\(\begin{array}{l}B=\frac{\mu _{0}I}{2\pi r}+\frac{\mu _{0}I}{4r}\end{array} \)

\(\begin{array}{l}B=\frac{\mu _{0}I}{4\pi r}(2+ \pi )\end{array} \)

Question 2) A conducting wire of length 24a is used to form two setups of coils, first in form of a square of side a and second in the form of an equilateral triangle of side a. Find the ratio of the magnetic moment of the coil in two cases.

(A) √3/1

(B) √2/1

\(\begin{array}{l}(C)\ \sqrt{\frac{3}{5}}\end{array} \)

\(\begin{array}{l}(D)\ \sqrt{\frac{5}{3}}\end{array} \)

Answer: (A)

Solution:

n₁ = Total length of the wire/Perimeter of the square

n₁ = 24a/4a

Thus, n₁= 6 turns

JEE Advanced Previous Year Questions With Answers on Magnetic Effect Of Current

n₂ = Total length of the wire/Perimeter of the triangle

3an₂ = 24a

⇒ n₂ = 8 turns

\(\begin{array}{l}\frac{M_{1}}{M_{2}}=\frac{n_{1}iA_{1}}{n_{2}iA_{2}}=\frac{6 \times a^{2}}{8\times a^{2}\times \frac{\sqrt{3}}{4}}= \frac{\sqrt{3}}{1}\end{array} \)

Question 3) An α -particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of a uniform magnetic field which is normal to the velocities of the particles. Within this region, the α-particle and the sulfur ion move in circular orbits of radii r_α and r_s , respectively. The ratio r_s/r_α is ________.

Answer (4)

Solution:

r = mv₀/qB

(½)mv₀² = qV

\(\begin{array}{l}r = \sqrt{2mqV}/qB\end{array} \)

\(\begin{array}{l}r = \frac{1}{B}\sqrt{\frac{2mV}{q}}\end{array} \)

\(\begin{array}{l}\frac{r_{s}}{r_{\alpha } }=\sqrt{\frac{m_{s}}{q_{s}}}\times\sqrt{\frac{q_{\alpha }}{m_{\alpha }}} =\sqrt{2}\times \sqrt{8}\end{array} \)

\(\begin{array}{l}\frac{r_{s}}{r_{\alpha } }= 4\end{array} \)

Question 4)

\(\begin{array}{l}\text{A physical quantity }\ \vec{S}\ \text{ is defined as}\ \vec{S} = (\vec{E}\times \vec{B})/\mu _{0},\end{array} \)

\(\begin{array}{l}\text{where}\ \vec{E}\ \text{ is electric field,}\ \vec{B}\end{array} \)
is magnetic field and μ₀ is the permeability of free space.
\(\begin{array}{l}\text{The dimensions of}\ \vec{S}\ \text{are the same as}\end{array} \)
the dimensions of which of the following quantity(ies)?

(A) Energy /(Charge x Current)

(B) Force/ (Length x Time)

(C) Energy/Volume

(D) Power/Area

Answer (B, D)

Solution:

\(\begin{array}{l}\vec{S} = (\vec{E}\times \vec{B})/\mu _{0}\end{array} \)

\(\begin{array}{l}\vec{S}\ \text{is known as Poynting vector and}\end{array} \)

represents intensity of electromagnetic waves

\(\begin{array}{l}\left [ \vec{S} \right ]=\left [ MT^{-3} \right ]=\left [ \frac{Power}{Area} \right ]= \left [ \frac{Force}{Length \times Time} \right ]\end{array} \)

Question 5) Two concentric circular loops, one of radius R and the other of radius 2R, lie in the xy-plane with the origin as their common center, as shown in the figure. The smaller loop carries current I₁in the anti-clockwise direction and the larger loop carries current I₂ in the clockwise direction, with I₂> 2I₁.

\(\begin{array}{l}\vec{B}(x,y)\ \text{denotes the magnetic field at a point (x, y) in the xy-plane.}\end{array} \)

Which of the following statement(s) is(are) correct?

\(\begin{array}{l}(A)\ \vec{B}(x,y)\ \text{is perpendicular to the xy-plane}\end{array} \)

at any point in the plane

\(\begin{array}{l}(B)\ \left | \vec{B}(x,y) \right |\ \text{depends on x and y only through the radial distance}\end{array} \)

\(\begin{array}{l}r =\sqrt{x^{2}+y^{2}}\end{array} \)

\(\begin{array}{l}(C)\ \left | \vec{B}(x,y) \right |\ \text{is non-zero at all points for}\ r < R\end{array} \)

\(\begin{array}{l}(D)\ \vec{B}(x,y)\ \text{points normally outward from the xy-plane}\end{array} \)

for all the points between the two loops

Answer (A, B)

Solution

A magnetic field due to a circular loop at any point in its plane will be perpendicular to the plane. Due to symmetry, it will depend only on the distance from the centre. The field will be in the opposite direction inside and outside the loop. The field may be non-zero for r < R, as it is in the opposite direction due to both the loops.

Question 6) A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius a, with its center at the origin. A magnetic dipole of moment m is brought along the axis of this loop from infinity to a point at distance r (>> a) from the center of the loop with its north pole always facing the loop, as shown in the figure below

The magnitude of magnetic field of a dipole m,

\(\begin{array}{l}\text{at a point on its axis at distance r, is }\ \frac{\mu _{0}m}{2\Pi r^{3}},\end{array} \)

where μ₀ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m₁ and m₂, separated by a distance r on the common axis, with their north poles facing each other, is km₁m₂/r⁴, where k is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

When the dipole m is placed at a distance r from the centre of the loop (as shown in the figure), the current induced in the loop will be proportional to

(A) m/r³

(B) m² /r²

(C) m/r²

(D) m²/r

Answer (A)

Solution

There will be induced emf which will oppose the change in flux

Flux, Φ = BA

Φ = (μ₀m/2πr³) x πa²

Induced emf, ε = |dΦ /dt|

ε = |(μ₀ma²/2) x (dr^-3/dt)|

= (3/2)(μ₀ma²/r⁴) v (since dr/dt = v)

Induced Current, I = ε/R

= (3/2)(μ₀ma²/r⁴R) v

I ∝ m/r⁴

Question 7) The dipole moment of a circular loop carrying current I is m and the magnetic field at the centre of the loop is B₁. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B₂. The ratio of B₁/B₂ is

(A) 2

(B) √3

(C) √2

(D) 1/√2

Answer: (C) √2

Solution

The magnetic field at the centre of the loop, B₁ = μ₀I/2R

Dipole moment of circular loop is m = IA

m₁ = I.A = I.πR² (R is the radius of the loop)

If the moment is doubled (keeping current constant) , R becomes √2R

JEE Advanced Previous Year Solved Paper on Magnetic Effect Of Current

Question 8)

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field B₁, at its centre when a current I passes through it. The second wire is bent into a coil with three identical loops adjacent to each other and produces magnetic field $B_{2}$ at the centre of the loops The ratio B₁: B₂ is

(A) 1:1

(B) 1:3

(C) 1:9

(D) 9:1

Answer: (B) 1:3

Solution

For loop B₁= μ₀I/2R₁

Where R₁ is the radius of the loop

2πR₁= l

⇒ R₁ = l/2π

Then, B₂ = 3μ₀(I/3)/2R₂

= μ₀I/2R₂

3 x 2πR₂= l

⇒ R₂ = l/6π

Therefore, B₁/B₂ = (R₂/R₁)

= (l/6π)/(l/2π) = 1: 3

Question 9) An electric current is flowing through a circular coil of radius R. The ratio of the magnetic field at the centre of the coil and that at a distance 2√2R from the centre of the coil and on its axis is:

(A) 2√2

(B) 27

(C) 36

(D) 8

Answer: (B) 27

Solution

Given, Radius = R

Distance, x = 2√2R

\(\begin{array}{l}\frac{B_{centre}}{B_{axis}}=\left [ 1+\frac{x^{2}}{R^{2}} \right ]^{3/2}=\left [ 1+\frac{(2\sqrt{2}R)^{2}}{R^{2}} \right ]^{3/2}=9^{3/2}=27\end{array} \)

Magnetic Effect of Current Jee Advanced Previous Year Questions with Solutions

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Magnetism JEE Main Previous Year Questions With Solutions

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