Magnetic Effect of Current — Page 9 | NEET Physics

Q81

A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance

{a \over 3}

and 2

a

, respectively from the axis of the wire is :

A 2

B

{1 \over 2}

C

{3 \over 2}

D

{2 \over 3}

Correct Answer

Option D

Solution

Let current density be J. $\therefore$ Applying Ampere's law. For point P

\oint {\overrightarrow {B.} d\overrightarrow l } = {\mu _0}i

$\Rightarrow$ BP2 $\pi$

{a \over 3}

=

{\mu _0}J\pi {{{a^2}} \over 9}

...(1) $\therefore$ Applying Ampere's law. For point Q BQ2 $\pi$

{(2a)}

=

{\mu _0}J\pi {a^2}

...(2) Dividing (1) by (2) we get

{{{B_P}2\pi {a \over 3}} \over {{B_Q}4\pi a}} = {{{\mu _0}J\pi {{{a^2}} \over 9}} \over {{\mu _0}J\pi {a^2}}}

$\Rightarrow$

{{{B_P}} \over {{B_Q}}} = {2 \over 3}

Q82

An electron is moving along the positive

\mathrm{x}

-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then A. The electron will experience magnetic force along positive y-axis B. The electron will experience magnetic force along negative y-axis C. The electron will not experience any force in magnetic field D. The electron will continue to move along the positive

\mathrm{x}

-axis E. The electron will move along circular path in magnetic field Choose the correct answer from the options given below:

A A and E only

B B and D only

C B and E only

D C and D only

Correct Answer

Option C

Solution

The Lorentz force equation is given as:

\vec{F} = -q(\vec{v} \times \vec{B})

The electron is moving along the positive x-axis, so its velocity vector is

\vec{v} = v_x \hat{i}

. The magnetic field is applied parallel to the negative z-axis, so its magnetic field vector is

\vec{B} = -B_z \hat{k}

. Now, we can calculate the cross product of the velocity and magnetic field vectors:

\vec{v} \times \vec{B} = (v_x \hat{i}) \times (-B_z \hat{k})

Using the cross product properties, we get:

\vec{v} \times \vec{B} = -v_x B_z (\hat{i} \times \hat{k})

The cross product of

\hat{i}

and

\hat{k}

is

-\hat{j}

, so:

\vec{v} \times \vec{B} = -v_x B_z (-\hat{j}) = v_x B_z \hat{j}

Since the electron has a negative charge, the magnetic force will be in the opposite direction:

\vec{F} = -(-e)(v_x B_z \hat{j}) = e(v_x B_z \hat{j})

As a result, the electron will experience a magnetic force along the negative y-axis.

Additionally, as mentioned earlier, when a charged particle moves through a magnetic field perpendicular to its velocity, it follows a circular path.

In this case, the velocity of the electron is along the positive x-axis, and the magnetic field is along the negative z-axis, which are indeed perpendicular to each other.

As a result, the electron will move along a circular path in the magnetic field.

Hence, the correct answer is: (C) B and E only

Q83

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be :

A [a/2, 3a]

B [a/4, 3a/2]

C [a/2, 2a]

D [a/4, 2a]

Correct Answer

Option C

Solution

Maximum possible magnetic field is at the surface

\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}

It can be obtained inside as well as outside the wire For inside,

\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \\ & \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2} \end{aligned}

For outside

\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \mathrm{r}=2 \mathrm{a} \end{aligned}

Correct answer $\left[\dfrac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$

Q84

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:

A

{{{\mu _0}IN} \over {2(b - a)}}{\log _e}\left( {{b \over a}} \right)

B

{{{\mu _0}I} \over 8}\left[ {{{a + b} \over {a - b}}} \right]

C

{{{\mu _0}I} \over {4(a - b)}}\left[ {{1 \over a} - {1 \over b}} \right]

D

{{{\mu _0}I} \over 8}\left( {{{a - b} \over {a + b}}} \right)

Correct Answer

Option A

Solution

No. of turns in dx width =

{N \over {b - a}}dx

\int {dB = \int\limits_a^b {\left( {{N \over {b - a}}} \right)dx{{{\mu _0}I} \over {2x}}} }

B = {{N{\mu _0}i} \over {2(b - a)}}\ln \left( {{b \over a}} \right)

Option (a)

Q85

A long straight wire of radius

a

carries a steady current

i.

The current is uniformly distributed across its cross section. The ratio of the magnetic field at

a/2

and

2a

is

A

1/2

B

1/4

C

4

D

1

Correct Answer

Option D

Solution

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the ampere-an path formed at a distance

{r_1}\left( { = {a \over 2}} \right)

= \left( {{{\pi r_1^2} \over {\pi {a^2}}}} \right) \times I,

where

I

is total current $\therefore$ Magnetic field at

{P_1}

is

{B_1} = {{{\mu _0} \times current\,\,enclosed} \over {path}}

\Rightarrow {B_1} = {{{\mu _0} \times \left( {{{\pi r_1^2} \over {\pi {a^2}}}} \right) \times I} \over {2\pi {r_1}}}

= {{{\mu _0} \times I{r_1}} \over {2\pi {a^2}}}

Now, magnetic field at point

{P_2},

{B_2} = {{{\mu _0}} \over {2\pi }}.{I \over {\left( {2a} \right)}} = {{{\mu _0}I} \over {4\pi a}}

$\therefore$ Required ratio

= {{{B_1}} \over {{B_2}}} = {{{\mu _0}I{r_1}} \over {2\pi {a^2}}} \times {{4\pi a} \over {{\mu _0}I}}

= {{2{r_1}} \over a} = {{2 \times {a \over 2}} \over a} = 1.

Q86

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path. Reason (R) : The magnetic field in that region is along the direction of velocity of the electron. In the light of the above statements, choose the correct answer from the options given below :

A (A) is true but (R) is false

B Both (A) and (R) are true and (R) is the correct explanation of (A)

C Both

(\mathbf{A})

and

(\mathbf{R})

are true but

(\mathbf{R})

is NOT the correct explanation of (A)

D (A) is false but (R) is true

Correct Answer

Option B

Solution

According to the Lorentz force law, the force experienced by a charged particle, such as an electron, moving in a magnetic field is given by the equation: $\overline{\mathrm{F}} = \mathrm{q}(\overline{\mathrm{v}} \times \overline{\mathrm{B}})$ Here, $\overline{\mathrm{F}}$ is the force exerted on the particle, $\mathrm{q}$ is the charge of the particle, $\overline{\mathrm{v}}$ is the velocity vector of the particle, and $\overline{\mathrm{B}}$ is the magnetic field vector.

The symbol $\times$ indicates the cross product, meaning the force is perpendicular to both the velocity and magnetic field vectors.

Given this, for the force $\overline{\mathrm{F}}$ to be zero: The velocity $\overline{\mathrm{v}}$ and the magnetic field $\overline{\mathrm{B}}$ must be parallel (or anti-parallel).

This condition will result in the angle $\theta = 0^\circ$ or $180^\circ$ .

Under these circumstances, the electron will not experience any magnetic force and will continue to move in a straight line with constant velocity, as the cross product becomes zero.

Therefore, the reason (R) is indeed a valid explanation: the magnetic field being along the direction of the electron's velocity means no perpendicular force is applied to change its path.

Q87

A coil of n number of turns wound tightly in the form of a spiral with inner and outer radii r1 and r2 respectively. When a current of strength I is passed through the coil, the magnetic field at its centre will be :

A

{{{\mu _0}nI} \over {2({r_2} - {r_1})}}

B

{{{\mu _0}nI} \over {{r_2}}}

C

{{{\mu _0}nI} \over {{r_2} - {r_1}}}{\log _e}{{{r_1}} \over {{r_2}}}

D

{{{\mu _0}nI} \over {2({r_2} - {r_1})}}{\log _e}{{{r_2}} \over {{r_1}}}

Correct Answer

Option D

Solution

In the width of

{r_2} - {r_1}

total n turns presents. $\therefore$ In 1 unit width

{n \over {{r_2} - {r_1}}}

turns presents. $\therefore$ In the width of dr number of turns,

n' = {n \over {{r_2} - {r_1}}} \times dr

Magnetic field (dB) due to element of dr length is

dB = {{{\mu _0} \times I \times n'} \over {2r}}

= {{{\mu _0}I} \over {2r}} \times {n \over {({r_2} - {r_1})}}

$\therefore$ Total magnetic field due to entire coil is,

\int {dB = \int_{{r_1}}^{{r_2}} {{{{\mu _0}I} \over {2r}} \times {n \over {({r_2} - {r_1})}}dr} }

\Rightarrow B = {{{\mu _0}In} \over {2({r_2} - {r_1})}}\int_{{r_1}}^{{r_2}} {{{dr} \over r}}

= {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times \left[ {\log _e^r} \right]_{{r_2}}^{{r_1}}

= {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times \left( {\log _e^{{r_2}} - \log _e^{{r_1}}} \right)

= {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times {\log _e}{{{r_2}} \over {{r_1}}}

Q88

A proton and a deutron

(q=+\mathrm{e}, m=2.0 \mathrm{u})

having same kinetic energies enter a region of uniform magnetic field

\vec{B}

, moving perpendicular to

\vec{B}

. The ratio of the radius

r_d

of deutron path to the radius

r_p

of the proton path is:

A

1: 2

B

1: 1

C

\sqrt{2}: 1

D

1: \sqrt{2}

Correct Answer

Option C

Solution

To solve for the ratio of the radii of deutron and proton paths in a magnetic field, we use the formula for the radius

r

of the circular path of a charged particle moving perpendicular to a uniform magnetic field:

r = \frac{mv}{qB}

where:

m

is the mass of the particle,

v

is the velocity of the particle,

q

is the charge of the particle, and

B

is the magnetic field strength. The proton and the deutron are given to have the same kinetic energy. The kinetic energy

K

of a particle is given by:

K = \frac{1}{2}mv^2

From the kinetic energy, we can express the velocity as:

v = \sqrt{\frac{2K}{m}}

Since both particles have the same kinetic energy and the charge of the deutron is the same as the charge of the proton

q = e

, but the mass of the deutron is twice that of the proton (

m_d = 2m_p

), substituting the expression for

v

in the radius formula, we get: For the deutron:

r_d = \frac{m_d\sqrt{2K/m_d}}{eB} = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_d}

For the proton (

m_p = m

):

r_p = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_p}

The ratio of the radius of the deutron path

r_d

to the radius of the proton path

r_p

is therefore:

\frac{r_d}{r_p} = \frac{\sqrt{m_d}}{\sqrt{m_p}} = \sqrt{\frac{2m_p}{m_p}} = \sqrt{2}

So, the correct answer is: Option C:

\sqrt{2}: 1

.

Q89

A long solenoid has

200

turns per

cm

and carries a current

i.

The magnetic field at its center is

6.28 \times {10^{ - 2}}\,\,\,Weber/{m^2}.

Another long solenoid has

100

turns per

cm

and it carries a current

{i \over 3}

. The value of the magnetic field at its center is

A

1.05 \times {10^{ - 2}}\,\,Weber/{m^2}

B

1.05 \times {10^{ - 5}}\,\,Weber/{m^2}

C

1.05 \times {10^{ - 3}}\,\,Weber/{m^2}

D

1.05 \times {10^{ - 4}}\,\,Weber/{m^2}

Correct Answer

Option A

Solution

{{{B_2}} \over {{B_1}}} = {{{\mu _0}{n_2}{i_2}} \over {{\mu _0}{n_1}{i_1}}}

\Rightarrow {{{B_2}} \over {6.28 \times {{10}^{ - 2}}}} = {{100 \times {i \over 3}} \over {200 \times i}}

\Rightarrow {B_2} = {{6.28 \times {{10}^{ - 2}}} \over 6}

= 1.05 \times {10^{ - 2}}\,\,Wb/{m^2}

Q90

To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle

\theta

. If a shunt of resistance S is needed to get half deflection then G, R and S are related by the equation :

A 2S (R + G) = RG

B S (R + G) = RG

C 2S = G

D 2G = S

Correct Answer

Option B

Solution

When only galvanometer G is present with the resistance R, Here IG =

{{{V_E}} \over {R + G}}

When shunt of resistance S is connected parallel to galvanometer, Here I =

{{{V_E}} \over {R + {{GS} \over {G + S}}}}

As deflection is half, here current through galvanometer, IG' =

{{{{\rm I}_G}} \over 2}

As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.

$\therefore$ IG' (G) = (I $-$ IG')S $\Rightarrow$ I'G (G + S) = IS $\Rightarrow$

{{{{\rm I}_G}} \over 2}

=

{{{\rm I}S} \over {G + S}}

$\Rightarrow$

{{{V_E}} \over {2\left( {R + G} \right)}}

=

{{{V_E}} \over {R + {{GS} \over {G + S}}}}

$\times$

{S \over {\left( {G + S} \right)}}

$\Rightarrow$

{1 \over {2\left( {R + G} \right)}}

=

{{G + S} \over {R(G + S) + GS}}

$\times$

{S \over {\left( {G + S} \right)}}

$\Rightarrow$ RG + RS + GS = 2RS + 2GS $\Rightarrow$ RG = RS + GS $\Rightarrow$ S(R + G) = RG