Magnetic Effect of Current — Page 8 | NEET Physics

Q71

To convert a galvanometer into a voltmeter one should connect a

A high resistance in series with galvanometer

B low resistance in series with galvanometer

C high resistance in parallel with galvanometer

D low resistance in parallel with galvanometer.

Correct Answer

Option A

Solution

For converting galvanometer to voltmeter, a high resistance should be connected in series with galvanometer.

Q72

An electron having mass m and kinetic energy E anter in uniform magnetic field B perpendiculaly, then its frequency will be

A

{{eE} \over {qvB}}

B

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

C

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

D

{{2m} \over {eBE}}

Correct Answer

Option C

Solution

The frequency of revolution of charged particle in a perpendicular magnetic field is

f = {1 \over T} = {1 \over {{{2\pi r} \over v}}} = {v \over {2\pi r}}

= {v \over {2\pi }} \times {{eB} \over {mv}} = {{eB} \over {2\pi m}}

Q73

If number of turns, area and current through a coil is given by n, A and

i

respectively then its magnetic moment will be

A

niA

B

{n^2}iA

C

ni{A^2}

D

{{ni} \over {\sqrt A }}

Correct Answer

Option A

Solution

Magnetic moment M = niA

Q74

The magnetic field at centre, P will be

A

{{{\mu _0}} \over {4\pi }}

B

{{{\mu _0}} \over {\pi }}

C

{{{\mu _0}} \over {2\pi }}

D

4{\mu _0}\pi

Correct Answer

Option C

Solution

Magnetic field due to 5A =

{{5{\mu _0}} \over {2\pi \times 2.5}}

=

{{2{\mu _0}} \over {2\pi }} \otimes

Magnetic field due to 2.5A =

{{2.5{\mu _0}} \over {2\pi \times 2.5}}

=

{{{\mu _0}} \over {2\pi }} \odot

Resultant Magnetic field =

{{2{\mu _0}} \over {2\pi }} - {{{\mu _0}} \over {2\pi }} = {{{\mu _0}} \over {2\pi }} \otimes

Q75

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicualr to the plane. Let rp, re and rHe be their respective radii, then

A re < rp < rHe

B re < rp = rHe

C re > rp > rHe

D re > rp = rHe

Correct Answer

Option B

Solution

r = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}

{r_{He}} = {r_p} > {r_e}

Q76

A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle :

A 1 : 1

B 1 : 2

C 2 : 1

D 1 : 4

Correct Answer

Option C

Solution

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

\Rightarrow R \propto \sqrt K

$\Rightarrow$ ratio = 2 : 1

Q77

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

A

{m \over \pi }

B

{{3m} \over \pi }

C

{{2m} \over \pi }

D

{{4m} \over \pi }

Correct Answer

Option D

Solution

Let the given square loop has side $a$ , then its magnetic dipole moment will be

m=I a^2

When square is converted into a circular loop of radius $r$ , Then, wire length will be same in both area,

\Rightarrow 4 a=2 \pi r \Rightarrow r=\frac{4 a}{2 \pi}=\frac{2 a}{\pi}

Hence, area of circular loop formed is, $A^{\prime}=\pi r^2$

=\pi\left(\frac{2 a}{\pi}\right)^2=\frac{4 a^2}{\pi}

Magnitude of magnetic dipole moment of circular loop will be

m^{\prime}=I A^{\prime}=I \frac{4 a^2}{\pi}

Ratio of magnetic dipole moments of both shapes is,

\begin{aligned} \frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \\\\ \Rightarrow m^{\prime} =\frac{4 m}{\pi}(\mathrm{A}-\mathrm{m}) \end{aligned}

Q78

The region between y = 0 and y = d contains a magnetic field

\overrightarrow B = B\widehat z

. A particle of mass m and charge q enters the region with a velocity

\overrightarrow v = v\widehat i.

If d

=

{{mv} \over {2qB}},

the acceleration of the charged particle at the point of its emergence at the other side is :

A

{{qvB} \over m}\left( -{{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)

B

{{qvB} \over m}\left( {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right)

C

{{qvB} \over m}\left( {{{ - \widehat j + \widehat i} \over {\sqrt 2 }}} \right)

D

{{qvB} \over m}\left( {{{\widehat j + \widehat i} \over {\sqrt 2 }}} \right)

Correct Answer

Option A

Solution

Here R =

{{mv} \over {qB}}

= 2d cos $\theta$ =

{{{R \over 2}} \over R}

=

{1 \over 2}

$\Rightarrow$ $\theta$ = 60o Acceleration of the charged particle at the point of its emergence,

\overrightarrow {{a_c}} = {a_{{c_x}}}\left( { - \widehat i} \right) + {a_{{c_y}}}\left( { - \widehat j} \right)

=

{a_c}\cos 30^\circ \left( { - \widehat i} \right) + {a_c}\sin 30^\circ \left( { - \widehat j} \right)

=

{a_c}\left( {{{\sqrt 3 } \over 2}\left( { - \widehat i} \right) + {1 \over 2}\left( { - \widehat j} \right)} \right)

=

{{qvB} \over m}\left( { - {{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)

Q79

A square loop of side 2

a

, and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about zaxis is given by :

A

{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}

B

{{{\mu _0}{I^2}{a^2}} \over {2\pi b}}

C

{{{\mu _0}{I^2}{a^3}} \over {2\pi {b^2}}}

D

{{2{\mu _0}{I^2}{a^3}} \over {\pi {b^2}}}

Correct Answer

Option A

Solution

We know,

\tau = MB\sin \theta

Here M = IA = I(2a)2 = 4a2I B =

{{{\mu _0}I} \over {2\pi b}}

Angle between B and M = 90o $\therefore$

\tau = \left( {4{a^2}I} \right)\left( {{{{\mu _0}I} \over {2\pi b}}} \right)\sin 90^\circ

=

{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}

Q80

A particle of mass m and charge q has an initial velocity

\overrightarrow v = {v_0}\widehat j

. If an electric field

\overrightarrow E = {E_0}\widehat i

and magnetic field

\overrightarrow B = {B_0}\widehat i

act on the particle, its speed will double after a time:

A

{{3m{v_0}} \over {q{E_0}}}

B

{{\sqrt 2 m{v_0}} \over {q{E_0}}}

C

{{\sqrt 3 m{v_0}} \over {q{E_0}}}

D

{{2m{v_0}} \over {q{E_0}}}

Correct Answer

Option C

Solution

Electric field will increase the speed of particle in x direction. Fx = qE $\therefore$ a =

{{qE} \over m}

Also vx = at =

{{qE} \over m}

t

v_x^2 + v_y^2 = {v^2}

$\Rightarrow$

v_x^2 + v_0^2 = {\left( {2{v_0}} \right)^2}

$\Rightarrow$ vx =

\sqrt 3

v0 $\therefore$

{{qE} \over m}

t =

\sqrt 3

v0 $\Rightarrow$ t =

{{\sqrt 3 m{v_0}} \over {q{E_0}}}