JEE Main Magnetic Effect of Current PYQ

46. (JEE Main 2019 (Online) 9th January Morning Slot )

An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d (d > > a). If the loop a applies a force F on the wire then :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Magnetic Effect of Current Question 121 English

A. F = 0

B. $F \propto (\frac{a}{d})$

C. $F \propto (\frac{a^{2}}{d^{3}})$

D. $F \propto {(\frac{a}{d})}^{2}$

Correct Answer is Option (D)

Magnetic field due to current carrying loop at distance d, is,

B = $\frac{μ_{0}}{4 π} \times \frac{\vec{M} \times \vec{d}}{d^{3}}$

= $\frac{μ_{0}}{4 π} \times \frac{i \times π a^{2} \times d \times \sin 90^{o}}{d^{3}}$    [as  M = iA]

= $\frac{μ_{0}}{4 π} \times \frac{i π a^{2}}{d^{2}}$

$∴$    B $\propto$ $\frac{a^{2}}{d^{2}}$

We also know,

F = Bil

$∴$    F $\propto$ B

So, F $\propto$ $\frac{a^{2}}{d^{2}}$

47. (JEE Main 2018 (Offline) )

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r_e, r_p, r $_{α}$ respectively in a uniform magnetic field B. The relation between r_e, r_p, r $_{α}$ is:

A. r_e < r $_{α}$ < r_p

B. r_e > r_p = r $_{α}$

C. r_e < r_p = r $_{α}$

D. r_e < r_p < r $_{α}$

Correct Answer is Option (C)

When a charged particle moves in a magnetic field then the charged particle moves in a circular path. So,

$\frac{m v^{2}}{r}$ = Bqv

$\Rightarrow$ r = $\frac{m v}{B q}$

We know kinetic energy, K = $\frac{1}{2}$ mv²

$∴$ mv = $\sqrt{2 K m}$

$∴$ r = $\frac{\sqrt{2 K m}}{B q}$

According to the question,

K_e (electron) = K_p (proton) = K_$α$(Alpha particle) = K = constant, and all of them are in uniform magnetic field.

$∴$ B = constant.

$∴$ r $\propto$ $\frac{\sqrt{m}}{q}$

For proton (₁H¹), mass = m, and charge = e

$∴$ r_p $\propto$ $\frac{\sqrt{m}}{e}$

For alpha particle (₂H⁴),

mass = 4m

and charge = 2e

$∴$ r_$α$ $\propto$ $\frac{\sqrt{4 m}}{2 e}$ $\propto$ $\frac{\sqrt{m}}{e}$

$∴$ r_p = r_$α$

For electron,

charge = e

and mass (m_e) = 9.1 $\times$ 10^{$-$ 31} kg

and mass of proton = 1.67 $\times$ 10^{$-$ 27} kg

$∴$ mass of electron < mass of proton.

r_e $\propto$ $\frac{\sqrt{m_{e}}}{e}$ < r_p

$∴$ r_e < r_p = r_$\propto$

48. (JEE Main 2017 (Online) 8th April Morning Slot )

In a certain region static electric and magnetic fields exist. The magnetic field is given by $\vec{B} = B_{0} (\hat{i} + 2 \hat{j} - 4 \hat{k})$ . If a test charge moving with a velocity $\vec{υ} = υ_{0} (3 \hat{i} - \hat{j} + 2 \hat{k})$ experiences no force in that region, then the electric field in the region, in SI units, is :

A. $\vec{E} = - υ_{0} B_{0} (3 \hat{i} - 2 \hat{j} - 4 \hat{k})$

B. $\vec{E} = - υ_{0} B_{0} (\hat{i} + \hat{j} + 7 \hat{k})$

C. $\vec{E} = υ_{0} B_{0} (14 \hat{j} + 7 \hat{k})$

D. $\vec{E} = - υ_{0} B_{0} (14 \hat{j} + 7 \hat{k})$

Correct Answer is Option (D)

Here test charge experience no net force, So, sum of electric and magnetic field is zero.

$∴$ F_e + F_m = 0

$∴$ F_e = $-$ q ( $\vec{v}$ $\times$ $\vec{B})$

= $-$ qB₀ $υ$ ₀ [(3 $\hat{i}$ $-$ $\hat{j}$ + 2 $\hat{k}$ ) $\times$ ( $\hat{i}$ + 2 $\hat{j}$ $-$ 4 $\hat{k}$ )]

= $-$ q $υ$ ₀ B₀ (14 $\hat{j}$ + 7 $\hat{k}$ )

Electric field produced by the charge q,

$\vec{E}$ = $\frac{\vec{F_{e}}}{q}$

= $\frac{- q υ_{0} B_{0} (14 \hat{j} + 7 \hat{k})}{q}$

= $-$ $υ$ ₀ B₀ (14 $\hat{j}$ + 7 $\hat{k}$ )

49. (AIEEE 2012 )

Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, denuteron and alpha particle are respectively $r_{p}, r_{d}$ and $r_{α}$ . Which one of the following relation is correct?

A. $r_{α} = r_{p} = r_{d}$

B. $r_{α} = r_{p} < r_{d}$

C. $r_{α} > r_{d} > r_{p}$

D. $r_{α} = r_{d} > r_{p}$

Correct Answer is Option (B)

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

Thus we have, $r_{α} = r_{p} < r_{d}$

50. (AIEEE 2007 )

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B},$ and comes out without any change in magnitude or direction of $\vec{v}$ . Then

A. $\vec{v} = \vec{B} \times \vec{E} / E^{2}$

B. $\vec{v} = \vec{E} \times \vec{B} / B^{2}$

C. $\vec{v} = \vec{B} \times \vec{E} / B^{2}$

D. $\vec{v} = \vec{E} \times \vec{B} / E^{2}$

Correct Answer is Option (B)

Here, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and the velocity $\vec{v}$ does not change; therefore

$q E = q v B \Rightarrow v = \frac{E}{B}$

Also, $| \frac{\vec{E} \times \vec{B}}{B^{2}} | = \frac{E B \sin θ}{B^{2}}$

$= \frac{E B \sin 90^{\circ}}{B^{2}} = \frac{E}{B} = | \vec{v} | = v$