JEE Main Magnetic Effect of Current PYQ

46. ( JEE Main 2021 (Online) 31st August Evening Shift)

A long solenoid with 1000 turns/m has a core material with relative permeability 500 and volume 10³ cm³. If the core material is replaced by another material having relative permeability of 750 with same volume maintaining same current of 0.75 A in the solenoid, the fractional change in the magnetic moment of the core would be approximately $(\frac{x}{499})$ . Find the value of x.

Correct Answer is 250

$\frac{Δ M}{M} = \frac{Δ μ}{μ} = \frac{250}{500} = \frac{1}{2}$

$\Rightarrow$ $\frac{1}{2} = \frac{x}{499} \Rightarrow x ≃ 250$

47. (JEE Main 2021 (Online) 27th August Morning Shift)

A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then in the form of a square of side 'a'. The coil is connected to a voltage source V₀. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1 : $\sqrt{y}$ where y is ................. .

Correct Answer is 3

In triangle shape $N_{t} = \frac{24 a}{3 a} = 8$

In square $N_{s} = \frac{24 a}{4 a} = 6$

$\frac{M_{t}}{M_{3}} = \frac{N_{t} I A_{t}}{N_{s} I A_{s}}$ [I will be same in both]

$= \frac{8 \times \frac{\sqrt{3}}{4} \times a^{2}}{6 \times a^{2}}$

$\frac{M_{t}}{M_{s}} = \frac{1}{\sqrt{3}}$

y = 3

48. (JEE Main 2020 (Online) 6th September Evening Slot )

A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field $\vec{B}$ . The magnetic moment of this moving particle:

A. $\frac{m v^{2} \vec{B}}{2 B^{2}}$

B. - $\frac{m v^{2} \vec{B}}{2 B^{2}}$

C. - $\frac{m v^{2} \vec{B}}{B^{2}}$

D. - $\frac{m v^{2} \vec{B}}{2 π B^{2}}$

Correct Answer is Option (B)

JEE Main 2020 (Online) 6th September Evening Slot Physics - Magnetic Effect of Current Question 79 English Explanation
Magnetic dipole moment

M = iA

$\Rightarrow$ M = $(\frac{q}{T}) \times π R^{2}$

= $\frac{q π R^{2}}{(\frac{2 π r}{v})}$ = $\frac{q v R}{2}$

$\Rightarrow$ M = $\frac{q v}{2} \times \frac{v m}{q B}$

As you can see from the figure, direction of magnetic moment(M) is opposite to magnetic field.

$∴$ $\vec{M} = - \frac{m v^{2}}{2 B} \hat{B}$

= $- \frac{m v^{2}}{2 B} (\frac{\vec{B}}{B})$

= $- \frac{m v^{2}}{2 B^{2}} \vec{B}$

49. (JEE Main 2020 (Online) 5th September Evening Slot )

An iron rod of volume 10^–3 m³ and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :

A. 5 $\times$ 10² Am²

B. 0.5 $\times$ 10² Am²

C. 500 $\times$ 10² Am²

D. 50 $\times$ 10² Am²

Correct Answer is Option (A)

Given, V = 10^–3 m³ = Al

I = 0.5A

$μ$ _r = 1000

n = 10 turns/cm = $\frac{10}{10^{- 2}}$ turn/m = 1000 turn/m

Magnetic moment, M = NIA( $μ$ _r - 1)

= (nl)IA( $μ$ _r - 1)

= nI(Al)( $μ$ _r - 1)

= 1000 × 0.5 × 10^–3 (1000 – 1)

= 0.5 × (999) = 499.5

$≃$ 500 = 5 $\times$ 10² Am²

50. (JEE Main 2020 (Online) 2nd September Evening Slot )

A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths a and b, then the magnitude and direction of magnetic moment of the loop ABCDEFA is JEE Main 2020 (Online) 2nd September Evening Slot Physics - Magnetic Effect of Current Question 88 English

A. $\sqrt{2}$ abI, along $(\frac{\hat{j}}{\sqrt{5}} + \frac{2 \hat{k}}{\sqrt{5}})$

B. abI, along $(\frac{\hat{j}}{\sqrt{5}} + \frac{2 \hat{k}}{\sqrt{5}})$

C. $\sqrt{2}$ abI, along $(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}})$

D. abI, along $(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}})$

Correct Answer is Option (C)

For ABCD

${\vec{M}}_{1}$ = abI $\hat{k}$

For DEFA

${\vec{M}}_{2}$ = abI $\hat{j}$

$\vec{M} = {\vec{M}}_{1} + {\vec{M}}_{2}$

= $a b I (\hat{k} + \hat{j})$

= $a b I \sqrt{2} (\frac{\hat{k}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}})$