Magnetic Effect of Current — Page 11 | JEE Physics

Q101

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

A

\sqrt{\dfrac{h}{\pi e B}}

B

\sqrt{\dfrac{4 h}{\pi e B}}

C

\sqrt{\dfrac{\mathrm{h}}{2 \pi e \mathrm{eB}}}

D

\sqrt{\dfrac{2 h}{\pi e B}}

Correct Answer

Option A

Solution

When an electron is projected perpendicular to a uniform magnetic field $B$ , it travels in a circular path.

According to Bohr's quantization rule, the radius of the electron's orbit in the first excited state can be determined using the following approach: The radius $r$ in a magnetic field is expressed as: $r = \dfrac{mv}{eB}$ where $m$ is the mass of the electron, $v$ is its velocity, $e$ is its charge, and $B$ is the magnetic field strength.

According to Bohr's quantization condition: $mvr = \dfrac{nh}{2\pi}$ Substituting the expression for $r$ : $(eBr)r = \dfrac{nh}{2\pi}$ This simplifies to the expression for the radius: $r = \sqrt{\dfrac{nh}{2\pi eB}}$ In the first excited state, the principal quantum number $n$ is 2.

So, the radius becomes: $r = \sqrt{\dfrac{2h}{2\pi eB}} = \sqrt{\dfrac{h}{\pi eB}}$ Thus, for the first excited state, the radius of the electron's orbit is $\sqrt{\dfrac{h}{\pi eB}}$ .

Q102

Let

B_1

be the magnitude of magnetic field at center of a circular coil of radius

R

carrying current I. Let

\mathrm{B}_2

be the magnitude of magnetic field at an axial distance '

x

' from the center. For

x: \mathrm{R}=3: 4, \dfrac{\mathrm{~B}_2}{\mathrm{~B}_1}

is :

A

64: 125

B

25: 16

C

4: 5

D

16: 25

Correct Answer

Option A

Solution

\begin{aligned} & B_1=\frac{\mu_0 i}{2 R} \quad \quad B_2=B_1 \sin ^3 \theta \\ & \therefore \frac{B_2}{B_1}=\sin ^3 \theta=\left(\frac{4}{5}\right)^3=\frac{64}{125} \end{aligned}

Q103

In a moving coil galvanometer, two moving coils

\mathrm{M}_1

and

\mathrm{M}_2

have the following particulars :

\begin{aligned} & \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_1=0.25 \mathrm{~T} \\ & \mathrm{R}_2=7 \Omega, \mathrm{~N}_2=21, \mathrm{~A}_2=1.8 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_2=0.50 \mathrm{~T} \end{aligned}

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of

M_1

and

M_2

?

A

1: 1

B

1: 3

C

1: 2

D

1: 4

Correct Answer

Option A

Solution

In a moving coil galvanometer, the voltage sensitivity is given by the formula: $\text{Voltage sensitivity} = \dfrac{\theta}{V} = \dfrac{N \cdot A \cdot B}{c \cdot R}$ Here, $\theta$ is the deflection angle, $V$ is the voltage, $N$ is the number of turns, $A$ is the coil area, $B$ is the magnetic field, $c$ is the torsional constant of the spring, and $R$ is the resistance of the coil.

For the two moving coils $\mathrm{M}_1$ and $\mathrm{M}_2$ , we want to determine the ratio of their voltage sensitivities.

Given that the torsional constant $c$ is the same for both coils, we can express the ratio as: $\text{Ratio of voltage sensitivities} = \left(\dfrac{N_1 \cdot A_1 \cdot B_1}{N_2 \cdot A_2 \cdot B_2}\right) \cdot \dfrac{R_2}{R_1}$ Plugging in the given values: $N_1 = 15$ $A_1 = 3.6 \times 10^{-3} \, \text{m}^2$ $B_1 = 0.25 \, \text{T}$ $R_1 = 5 \, \Omega$ $N_2 = 21$ $A_2 = 1.8 \times 10^{-3} \, \text{m}^2$ $B_2 = 0.50 \, \text{T}$ $R_2 = 7 \, \Omega$ We calculate: $\text{Ratio} = \left(\dfrac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5}\right) \times \dfrac{7}{5} = \dfrac{1}{1}$ Thus, the ratio of the voltage sensitivities of $M_1$ to $M_2$ is 1:1.

Q104

A solenoid having area A and length '

l

' is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

A

\dfrac{\mathrm{B}^2 \mathrm{~A} l}{4 \mu_0}

B

\dfrac{\mathrm{B}^2 \mathrm{~A} l}{2 \mu_0}

C

\dfrac{\mathrm{B}^2 \mathrm{~A} l}{\mu_0}

D

\mathrm{B}^2 \mathrm{Al}

Correct Answer

Option A

Solution

To determine the magnetic energy stored in a solenoid filled with a material of relative permeability 2, we start with the expression for energy density in a magnetic field: $\dfrac{U}{V} = \dfrac{B^2}{2 \mu_{\mathrm{r}} \mu_0}$ Given that the relative permeability $\mu_{\mathrm{r}}$ is 2, this becomes: $\dfrac{U}{V} = \dfrac{B^2}{2 \times 2 \mu_0} = \dfrac{B^2}{4 \mu_0}$ The energy $U$ stored in the solenoid can be expressed as: $U = \dfrac{B^2}{4 \mu_0} \times V$ Here, $V$ is the volume of the solenoid, calculated as $A \times \ell$ (where $A$ is the cross-sectional area and $\ell$ is the length).

Therefore, substituting for $V$ , we get: $U = \dfrac{B^2}{4 \mu_0} \times A \ell$ This formula gives us the magnetic energy stored in the solenoid.

Q105

The dipole moment of a circular loop carrying a 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

{{B_2}}

. The ratio

{{{B_1}} \over {{B_2}}}

is:

A 2

B

\sqrt 3

C

\sqrt 2

D

1 \over \sqrt 2

Correct Answer

Option C

Solution

Dipole moment, M = IA Let radius of circular loop = R

\therefore\,\,\,

M = I $\times$ $\pi$ R2 Later, we keep current constant , But dipole moment becomes double, let new radius = R1

\therefore\,\,\,

2M = I $\times$ $\pi$ R

_1^2

$\Rightarrow$

\,\,\,

2I $\pi$ R2 = I $\pi$ R

_1^2

$\Rightarrow$

\,\,\,

R1 =

\sqrt 2

R At the center of circular ring, the magnetic field, B =

{{{\mu _0}I} \over {2R}}

\therefore\,\,\,

B1 =

{{{\mu _0}I} \over {2R}}

and B2 =

{{{\mu _0}I} \over {2 \times \left( {\sqrt 2 R} \right)}}

\therefore\,\,\,

{{{B_1}} \over {{B_2}}}

=

{{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}

=

{{2\sqrt 2 } \over 2}

=

\sqrt 2

Q106

Given below are two statements: one is labelled as

\mathbf{A s s e r t i o n} \mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : If Oxygen ion

\left(\mathrm{O}^{-2}\right)

and Hydrogen ion

\left(\mathrm{H}^{+}\right)

enter normal to the magnetic field with equal momentum, then the path of

\mathrm{O}^{-2}

ion has a smaller curvature than that of

\mathrm{H}^{+}

. Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. In the light of the above statements, choose the correct answer from the options given below

A Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

B

\mathbf{A}

is false but

\mathbf{R}

is true

C

\mathbf{A}

is true but

\mathbf{R}

is false

D Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

Correct Answer

Option C

Solution

The explanation involves analyzing the motion of charged particles in a magnetic field using the concept of radius of curvature.

When a charged particle moves perpendicularly to a magnetic field, the radius of curvature $r$ of its path is given by: $r = \dfrac{mv}{qB} = \dfrac{p}{qB}$ Where: $m$ is the mass of the particle. $v$ is the velocity. $q$ is the charge. $B$ is the magnetic field strength. $p$ is the momentum of the particle.

From this formula, we can see that the radius of curvature $r$ is directly proportional to the momentum $p$ and inversely proportional to the charge $q$ .

In the given scenario: The Oxygen ion $\mathrm{O}^{-2}$ and the Hydrogen ion $\mathrm{H}^{+}$ both enter the magnetic field with equal momentum.

Since $\mathrm{O}^{-2}$ has a charge of $-2e$ and $\mathrm{H}^{+}$ has a charge of $+e$ , and given that $r \propto \dfrac{1}{q}$ , the path of $\mathrm{O}^{-2}$ will indeed have a smaller radius of curvature compared to that of $\mathrm{H}^{+}$ .

Therefore, the Assertion $\mathbf{A}$ is true.

The Reason $\mathbf{R}$ states that a proton and an electron with the same linear momentum will form paths of different radii upon entering a magnetic field, with the proton having a smaller radius of curvature.

This reasoning is based on the same principle, since the mass and charge of the electron and proton are different.

However, this explanation does not directly justify the assertion about Oxygen and Hydrogen ions, so Reason $\mathbf{R}$ is false in the context of explaining Assertion $\mathbf{A}$ .

Thus, the correct conclusion is that $\mathbf{A}$ is true, but $\mathbf{R}$ is false.

Q107

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the radio of the magnetic field at the central of the loop (BL) to that at the center of the coil (BC), i.e.

{{{B_L}} \over {{B_C}}}

will be :

A N

B

{1 \over N}

C N2

D

{1 \over {{N^2}}}

Correct Answer

Option D

Solution

For loop, L = 2 $\pi$ R For coil, L = N $\times$ 2 $\pi$ r $\therefore$ 2 $\pi$ R = N $\times$ 2 $\pi$ r $\Rightarrow$ R = Nr $\Rightarrow$ r =

{R \over N}

We know, BL =

{{{\mu _0}i} \over {2R}}

and BC = N $\times$

{{{\mu _0}i} \over {2r}}

$\therefore$

{{{B_L}} \over {{B_C}}} = {{{{{\mu _0}i} \over {2R}}} \over {N \times {{{\mu _0}i} \over {2\left( {{R \over N}} \right)}}}} = {1 \over {{N^2}}}

Q108

An insulating thin rod of length

l

has a linear charge density

\rho \left( x \right)

=

{\rho _0}{x \over l}

on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is -

A

{\pi \over 3}n\rho {l^3}

B

{\pi \over 4}n\rho {l^3}

C

n\rho {l^3}

D

\pi n\rho {l^3}

Correct Answer

Option B

Solution

$\because$ M = NIA dq = $\lambda$ dx & A = $\pi$ x2

\int {dm} = \int {\left( x \right){{{\rho _0}x} \over \ell }} \,dx.\pi {x^2}

M =

{{n{\rho _0}\pi } \over \ell }.\int\limits_0^\ell {{x^3}.dx} = {{n{\rho _0}\pi } \over \ell }.\left[ {{{{L^4}} \over 4}} \right]

M =

{{n{\rho _0}\pi {\ell ^3}} \over 4}

or

{\pi \over 4}n\rho {\ell ^3}

Q109

In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6

\times

10–19 C Mass of the electron = 9.1

\times

10–31 kg]

A 7.5

\times

10

-

4 m

B 7.5

\times

10

-

3 m

C 7.5 m

D 7.5

\times

10

-

2 m

Correct Answer

Option A

Solution

r = {{\sqrt {2mk} } \over {eB}} = {{\sqrt {2me\Delta v} } \over {eB}}

r = {{\sqrt {{{2m} \over e}.\Delta v} } \over B} = {{\sqrt {{{2 \times 9.1 \times {{10}^{ - 31}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {500} \right)} } \over {100 \times {{10}^{ - 3}}}}

r = {{\sqrt {{{9.1} \over {0.16}} \times {{10}^{ - 10}}} } \over {{{10}^{ - 1}}}} = {3 \over 4} \times {10^{ - 4}}

= 7.5 \times {10^{ - 4}}

Q110

A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75o. One of the fields has a magnitude of 15 mT. The dipole attains stable equilibrium at an angle of 30o with this field. The magnitude of the other field (in mT ) is close to

A 11

B 36

C 1

D 1060

Correct Answer

Option A

Solution

For equilibrium, net torque acting on dipole is = 0 $\therefore$ $\tau$ 1 = $\tau$ 2 $\Rightarrow$ mB1 sin $\theta$ 1 = mB2 sin $\theta$ 2 $\Rightarrow$ B2 = B1 $\times$

{{\sin {{30}^o}} \over {\sin {{45}^o}}}

= 15 $\times$

\sqrt 2

$\times$

{1 \over 2}

= = 10.6 mT

\simeq

11 mT