Practice Start Test

Magnetic Effect of Current

NEET Physics · 96 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
A metallic rod of mass per unit length 0.5 kg m –1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
A 7.14 A
B 5.98 A
C 14.76 A
D 11.32 A
Correct Answer
Option D
Solution

Given Mass per unit length of a metallic rod is

ml{m \over l}

= 0.5 kg m -1 From figure, for equilibrium mg sin 30 o = I

ll

B cos 30 o ⇒\Rightarrow I =

mglBtan⁡30∘{{mg} \over {lB}}\tan 30^\circ

=

0.5×9.80.25×3{{0.5 \times 9.8} \over {0.25 \times \sqrt 3 }}

= 11.32 A

Q22
Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
A 40 Ω\Omega
B 25 Ω\Omega
C 250 Ω\Omega
D 500 Ω\Omega
Correct Answer
Option C
Solution

Current sensitivity of moving coil galvanometer I s =

NBAC{{NBA} \over C}

.....(i) Voltage sensitivity of moving coil galvanometer, V s =

NBACRG{{NBA} \over {C{R_G}}}

Dividing eqn. (i) by (ii) Resistance of galvanometer R G =

IsVS=5×120×10−3=500020=250 Ω{{{I_s}} \over {{V_S}}} = {{5 \times 1} \over {20 \times {{10}^{ - 3}}}} = {{5000} \over {20}} = 250\,\Omega
Q23
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ′I′'I' along the same direction as shown in fogure. Magnitude of force per unit length on the middle wire ′B′'B' is given by
A 2μ0I2πd{{2{\mu _0}{I^2}} \over {\pi d}}
B 2μ0I2πd{{\sqrt 2 {\mu _0}{I^2}} \over {\pi d}}
C μ0I22πd{{{\mu _0}{I^2}} \over {\sqrt 2 \pi d}}
D μ0I22πd{{{\mu _0}{I^2}} \over {2\pi d}}
Correct Answer
Option C
Solution

Since same current flowing through both the wires i i = i 2 = i So F 1 =

μ0i22πd{{{\mu _0}{i^2}} \over {2\pi d}}

= F 2 ∴\therefore Magnitude of force per unit length on the middle wire 'B' F net =

F12+F22\sqrt {F_1^2 + F_2^2}

=

μ0i22πd{{{\mu _0}{i^2}} \over {\sqrt 2 \pi d}}
Q24
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × \times 10 -2 T. If the value of e/m is 1.76 × \times 10 11 C kg −-1 , the frequency of revoluation of the electron is
A 1 GHz
B 100 MHz
C 62.8 MHz
D 6.28 MHz
Correct Answer
Option A
Solution

Frequency of revolution of charge in magnetic field is given as F =

qB2πm{{qB} \over {2\pi m}}

=

(1.76×1011)×3.57×10−22×3.14{{\left( {1.76 \times {{10}^{11}}} \right) \times 3.57 \times {{10}^{ - 2}}} \over {2 \times 3.14}}

= 10 9 Hz = 1 GHz

Q25
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
A nB
B n 2 B
C 2nB
D 2n 2 B
Correct Answer
Option B
Solution

For One turn loop : B =

μ0i2r{{{\mu _0}i} \over {2r}}

For n turn loop : r =

Rn{R \over n}

B' =

μ0ni2Rn{{{\mu _0}ni} \over {2{R \over n}}}

=

(μ0i2R)n2\left( {{{{\mu _0}i} \over {2R}}} \right){n^2}

= n 2 B

Q26
A square loop ABCD carrying a current ii, is placed near and coplanar with a long straight conductor XY carrying a current II, the net force on the loop will be
A 2μ0liL3π{{2{\mu _0}liL} \over {3\pi }}
B μ0liL2π{{{\mu _0}liL} \over {2\pi }}
C 2μ0li3π{{2{\mu _0}li} \over {3\pi }}
D μ0li2π{{{\mu _0}li} \over {2\pi }}
Correct Answer
Option C
Solution

Force on arm AB due to current in conductor XY is

F1=μ04π2IiL(L/2)=μ0Iiπ{F_1} = {{{\mu _0}} \over {4\pi }}{{2IiL} \over {\left( {L/2} \right)}} = {{{\mu _0}Ii} \over \pi }

acting towards XY in the plane of loop. Force on arm CD due to current in conductor XY is

F2=μ04π2IiL3(L/2)=μ0Ii3π{F_2} = {{{\mu _0}} \over {4\pi }}{{2IiL} \over {3\left( {L/2} \right)}} = {{{\mu _0}Ii} \over {3\pi }}

acting away from XY in the plane of loop. ∴\therefore Net force on the loop = F 1 – F 2

=μ0Ii3π[1−13]=23μ0Iiπ= {{{\mu _0}Ii} \over {3\pi }}\left[ {1 - {1 \over 3}} \right] = {2 \over 3}{{{\mu _0}Ii} \over \pi }
Q27
A long straight wire of radius aa carries a steady current II. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distance a2{a \over 2} and 2aa respectively, from the axis of the wire is
A 1
B 4
C 14{1 \over 4}
D 12{1 \over 2}
Correct Answer
Option A
Solution

For points inside the wire i.e., (r ≤\le R) Magnetic field

B=μ0Ir2πR2B = {{{\mu _0}Ir} \over {2\pi {R^2}}}

For points outside the wire (r ≥\ge R) Magnetic field,

B′=μ0I2πRB' = {{{\mu _0}I} \over {2\pi R}}
∴BB′=μ0I(a/2)2πa2μ0I2π(2a)=1:1\therefore {B \over {B'}} = {{{{{\mu _0}I\left( {a/2} \right)} \over {2\pi {a^2}}}} \over {{{{\mu _0}I} \over {2\pi \left( {2a} \right)}}}} = 1:1
Q28
A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be
A 1.5 MeV
B 1 MeV
C 4 MeV
D 0.5 MeV
Correct Answer
Option B
Solution

As we know, F = qvB =

mv2R{{m{v^2}} \over R}

∴\therefore

R=mvqB=2m(kE)qBR = {{mv} \over {qB}} = {{\sqrt {2m\left( {kE} \right)} } \over {qB}}

Since R is same so,

KE∝q2mKE \propto {{{q^2}} \over m}

Therefore KE of α\alpha particle

=q2m=(2)24=1 MeV= {{{q^2}} \over m} = {{{{\left( 2 \right)}^2}} \over 4} = 1\,MeV
Q29
A wire carrying current II has the shape shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular protion of radius R is lying in Y-Z plane. Magtnetic field at pont OO is
A B→=−μ0I4πR(πi^+2k^)\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)
B B→=μ0I4πR(πi^−2k^)\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right)
C B→=μ0I4πR(πi^+2k^)\overrightarrow B = {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)
D B→=−μ0I4πR(πi^−2k^)\overrightarrow B = - {{{\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i - 2\widehat k} \right)
Correct Answer
Option A
Solution

Magnetic field due to segment ‘1’

B1→=μ0I4πR[sin⁡90∘+sin⁡0∘](−k^)\overrightarrow {{B_1}} = {{{\mu _0}I} \over {4\pi R}}\left[ {\sin 90^\circ + \sin 0^\circ } \right]\left( { - \widehat k} \right)
=−μ0I4πR(k^)=B3→= {{ - {\mu _0}I} \over {4\pi R}}\left( {\widehat k} \right) = \overrightarrow {{B_3}}

Magnetic field due to segment 2

B2→=μ0I4R(−i^)=−μ0I4πR(−πi^)\overrightarrow {{B_2}} = {{{\mu _0}I} \over {4R}}\left( { - \widehat i} \right) = {{ - {\mu _0}I} \over {4\pi R}}\left( { - \pi \widehat i} \right)
∴B→\therefore \overrightarrow B

at centre

Bc→=B1→+B2→+B3→\overrightarrow {{B_c}} = \overrightarrow {{B_1}} + \overrightarrow {{B_2}} + \overrightarrow {{B_3}}
=−μ0I4πR(πi^+2k^)= {{ - {\mu _0}I} \over {4\pi R}}\left( {\pi \widehat i + 2\widehat k} \right)
Q30
A conducting square frame of side 'a' and a long straight wire carrying current II are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity 'V'. The emf induced in the frame will be proportional to
A 1(2x+a)2{1 \over {{{\left( {2x + a} \right)}^2}}}
B 1(2x−a)(2x+a){1 \over {\left( {2x - a} \right)\left( {2x + a} \right)}}
C 1x2{1 \over {{x^2}}}
D 1(2x−a)2{1 \over {{{\left( {2x - a} \right)}^2}}}
Correct Answer
Option B
Solution

The emf in AD: e 1 = (a × μ 0 iv)/2π(x −a/2) The emf in EF : e 2 = (a × μ 0 i × v)/2π(x + a/2) Net emf = e 1 – e 2

⇒e=μ02πa×i×v[1x−a2−1x+a2]\Rightarrow e = {{{\mu _0}} \over {2\pi }}a \times i \times v\left[ {{1 \over {x - {a \over 2}}} - {1 \over {x + {a \over 2}}}} \right]
e∝1(2x−a)(2x+a)e \propto {1 \over {\left( {2x - a} \right)\left( {2x + a} \right)}}
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