Magnetic Effect of Current — Page 12 | JEE Physics

Q111

A particle having the same charge as of electron moves in a ciurcular path of radius 0.5 cm under the influence of a magnetic field 0f 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6

\times

10

-

19C)

A 9.1

\times

10

-

31 kg

B 1.6

\times

10

-

27 kg

C 1.6

\times

10

-

19 kg

D 2.0

\times

10

-

24 kg

Correct Answer

Option D

Solution

Given, radius of circular path(r) = 0.5 cm Magnetic field (B) = 0.5 T Electric field (E) = 100 V/m Charge of particle (q) = 1.6 $\times$ 10 $-$ 19 C As particle is moving in a circular path so,

{{m{v^2}} \over r} = qvB

$\Rightarrow$ r =

{{mv} \over {qB}}

. . . . . . .

(1) When electric field of 100 v/m is applied on the particle then particle is moving in the straight line.

So, the net force on the particle is zero.

$\therefore$ Fnet = 0 $\Rightarrow$ Fe = Fm $\Rightarrow$ qE = qvB $\Rightarrow$ E = vB . . . . .(

2) From equation (1) and (2) we get, r =

{m \over {qB}}\left( {{E \over B}} \right)

=

{{mE} \over {q{B^2}}}

$\Rightarrow$ m =

{{q{B^2}r} \over E}

=

{{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}

= 2 $\times$ 10 $-$ 24 kg

Q112

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)

Q113

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.

Q114

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.

Q115

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}}}

Q116

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

.

Q117

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}

Q118

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

Q119

A horizontal overhead powerline is at height of

4m

from the ground and carries a current of

100A

from east to west. The magnetic field directly below it on the ground is

\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)

A

2.5 \times {10^{ - 7}}\,T

southward

B

5 \times {10^{ - 6}}\,T

northward

C

5 \times {10^{ - 6}}\,T

southward

D

2.5 \times {10^{ - 7}}\,T

northward

Correct Answer

Option C

Solution

The magnetic field is

B = {{{\mu _0}} \over {4\pi }}{{2I} \over r}

= {10^{ - 7}} \times {{2 \times 100} \over 4}

= 5 \times {10^{ - 6}}T

According to right hand palm rule, the magnetic field is directed towards south.

Q120

A long solenoid is formed by winding 70 turns cm

^{-1}

. If 2.0 A current flows, then the magnetic field produced inside the solenoid is ____________ (

\mu_0=4\pi\times10^{-7}

TmA

^{-1}

)

A

88\times10^{-4}

T

B

1232\times10^{-4}

T

C

176\times10^{-4}

T

D

352\times10^{-4}

T

Correct Answer

Option C

Solution

Number of turns per meter $=7000$ turns per $\mathrm{m}$

\begin{aligned} &i=2 \mathrm{~A} & \\\\ & B=\mu_{0} n i =4 \pi \times 10^{-7} \times 7000 \times 2 \\\\ & =56 \pi \times 10^{-4} \mathrm{~T} \\\\ & =56 \times \frac{22}{7} \times 10^{-4} \mathrm{~T} \\\\ & =176 \times 10^{-4} \mathrm{~T} \end{aligned}