Magnetic Effect of Current — Page 8 | JEE Physics

Q71

Given below are two statements : Statement I : The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle. Statement II : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are correct.

B Both Statement I and Statement II are incorrect.

C Statement I is correct but Statement II is incorrect.

D Statement I is incorrect but Statement II is correct.

Correct Answer

Option C

Solution

Electric field accelerates the particle in the direction of field

\left( {\overrightarrow F = q\overrightarrow E = m\overrightarrow a } \right)

and magnetic field accelerates the particle perpendicular to the field

\left( {\overrightarrow F = q\overrightarrow v \times \overrightarrow B = m\overrightarrow a } \right)

.

Q72

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2

\times

10

-

6 N, then the value of x is approximately :

A 1

B 2.4

C 1.4

D 2

Correct Answer

Option C

Solution

{{dF} \over {dl}} = 2 \times {10^{ - 6}}

N/m

= {{{\mu _0}{i_1}{i_2}} \over {2\pi d}}

2 \times {10^{ - 6}} = {{2 \times {{10}^{ - 7}} \times {x^2}} \over {0.2}}

x = \sqrt 2 \simeq 1.4

Q73

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is :

A 1 : 4

B 4 : 1

C 2 : 1

D 1 : 2

Correct Answer

Option C

Solution

$R=\dfrac{m v}{q B}$ $\dfrac{\mathrm{R}_\alpha}{\mathrm{R}_{\mathrm{P}}}=\dfrac{\mathrm{M}_\alpha}{\mathrm{M}_{\mathrm{P}}} \times \dfrac{\mathrm{q}_{\mathrm{P}}}{\mathrm{q}_\alpha}$ $\dfrac{\mathrm{R}_\alpha}{\mathrm{R}_{\mathrm{P}}}=\dfrac{4}{1} \times \dfrac{1}{2}=2$

Q74

A proton, a deutron and an

\alpha

-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :

A 1 :

\sqrt 2

:

\sqrt 2

B 1 : 1 :

\sqrt 2

C

\sqrt 2

: 1 : 1

D 1 :

\sqrt 2

: 1

Correct Answer

Option D

Solution

$\therefore$

r = {{mv} \over {qB}} = {{\sqrt {2m(KE)} } \over {qB}}

\Rightarrow {r_1}:{r_2}:{r_3} = {{\sqrt {{m_1}} } \over {{q_1}}}:{{\sqrt {{m_2}} } \over {{q_2}}}:{{\sqrt {{m_3}} } \over {{q_3}}}

= {{\sqrt 1 } \over 1}:{{\sqrt 2 } \over 1}:{{\sqrt 4 } \over 2}

= 1:\sqrt 2 :1

Q75

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an uniform magnetic field, speed and energy remains the same for a moving charged particle. Reason (R) : Moving charged particle experiences magnetic force perpendicular to its direction of motion.

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

B Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

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

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

Correct Answer

Option A

Solution

Magnetic force

\overrightarrow F \bot \overrightarrow v

\Rightarrow {W_b} = 0

\Rightarrow \Delta KE = 0

and speed remains constant.

Q76

A cyclotron is working at a frequency of 10 MHz. If the radius of its dees is 60 cm. The maximum kinetic energy of accelerated proton will be : (Take : e = 1.6

\times

10

-

19 C, mp = 1.67

\times

10

-

27 kg)

A 7.4 MeV

B 14.86 MeV

C 7.4 GeV

D 704 GeV

Correct Answer

Option A

Solution

Given,

f = 10 \times {10^6}

Hz

r = 0.6

m Charge on proton (q) = e We know, Radius

(r) = {{mv} \over {qB}}

= {{\sqrt {2mk} } \over {eB}}

.....

(1) Also, we know, Cyclotron oscillation frequency should be equal to the pendulum evolution frequency.

$\therefore$

f = {{eB} \over {2\pi m}}

\Rightarrow eB = 2\pi mf

Putting this value of eB in equation (1) we get

r = {{\sqrt {2mk} } \over {2\pi mf}}

\Rightarrow {r^2} = {{2mk} \over {4{\pi ^2}{m^2}{f^2}}}

\Rightarrow k = 2{\pi ^2}m{f^2}{r^2}

= 2 \times {\left( {{{22} \over 7}} \right)^2} \times 1.67 \times {10^{ - 27}} \times {\left( {10 \times {{10}^6}} \right)^2} \times {\left( {0.6} \right)^2}

J

= 1.2 \times {10^{ - 12}}

J

= {{1.2 \times {{10}^{ - 12}}} \over {1.6 \times {{10}^{ - 19}}}}

eV

= {{12} \over {16}} \times {10^7}

eV

= 0.75 \times {10^7}

eV

= 7.5 \times {10^6}

eV = 7.5 MeV

Q77

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is

6: 5

and their respective masses ratio is

9: 4

. Then, the ratio of their charges will be :

A 8 : 5

B 5 : 4

C 5 : 3

D 8 : 7

Correct Answer

Option B

Solution

We know that

R = {{mv} \over {Bq}} = \sqrt {{{2mK} \over {Bq}}}

$\Rightarrow$ Ratio of radii

= {{{R_1}} \over {{R_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} {{{q_2}} \over {{q_1}}}

\Rightarrow {6 \over 5} = \sqrt {{9 \over 4}} {{{q_2}} \over {{q_1}}}

\Rightarrow {{{q_1}} \over {{q_2}}} = {3 \over 2} \times {5 \over 6} = {5 \over 4}

Q78

A charge particle is moving in a uniform magnetic field

(2 \hat{i}+3 \hat{j}) \,\mathrm{T}

. If it has an acceleration of

(\alpha \hat{i}-4 \hat{j})\, \mathrm{m} / \mathrm{s}^{2}

, then the value of

\alpha

will be :

A 3

B 6

C 12

D 2

Correct Answer

Option B

Solution

As magnetic force is perpendicular to magnetic field So,

\overrightarrow F

.

\overrightarrow B

must be 0 So, 2 $\alpha$ $-$ 12 = 0 $\alpha$ = 6

Q79

\mathrm{B}_{X}

and

\mathrm{B}_{\mathrm{Y}}

are the magnetic fields at the centre of two coils

\mathrm{X}

and

\mathrm{Y}

respectively each carrying equal current. If coil

X

has 200 turns and

20 \mathrm{~cm}

radius and coil

Y

has 400 turns and

20 \mathrm{~cm}

radius, the ratio of

B_{X}

and

B_{Y}

is :

A 1 : 1

B 1 : 2

C 2 : 1

D 4 : 1

Correct Answer

Option B

Solution

B = {{{\mu _0}NI} \over {2R}}

{{{B_X}} \over {{B_Y}}} = {{{N_x}{R_y}} \over {{N_y}{R_x}}}

= {{200 \times 20} \over {400 \times 20}} = {1 \over 2}

Q80

A cyclotron is used to accelerate protons. If the operating magnetic field is

1.0 \mathrm{~T}

and the radius of the cyclotron 'dees' is

60 \mathrm{~cm}

, the kinetic energy of the accelerated protons in MeV will be :

[\mathrm{use} \,\,\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \,\mathrm{C}

]

A 12

B 18

C 16

D 32

Correct Answer

Option B

Solution

R = {{mv} \over {Bq}} = {{\sqrt {2mK} } \over {Bq}}

\Rightarrow K = {{{B^2}{q^2}{R^2}} \over {2m}}

= {{{{(1.6 \times {{10}^{ - 19}})}^2} \times {{0.6}^2}} \over {2 \times 1.6 \times {{10}^{ - 27}}}}

J

= 18

MeV