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Magnetic Effect of Current

NEET Physics · 96 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude
A μ0n2er{{{\mu _0}{n^2}e} \over r}
B μ0ne2r{{{\mu _0}ne} \over {2r}}
C μ0ne2πr{{{\mu _0}ne} \over {2\pi r}}
D Zero
Correct Answer
Option B
Solution

Current in the orbit,

I=eTI = {e \over T}
I=e(2π/ω)=ωe2πI = {e \over {\left( {2\pi /\omega } \right)}} = {{\omega e} \over {2\pi }}
=(2πn)e2π=ne= {{\left( {2\pi n} \right)e} \over {2\pi }} = ne

Magnetic field at centre of current carrying circular coil is given by

B=μ0I2r=μ0ne2rB = {{{\mu _0}I} \over {2r}} = {{{\mu _0}ne} \over {2r}}
Q32
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be
A 1499G{1 \over {499}}G
B 499500G{{499} \over {500}}G
C 1500G{1 \over {500}}G
D 500499G{{500} \over {499}}G
Correct Answer
Option C
Solution

In parallel arrangement, I α\alpha R Now G/S = 99.8/0.2 S = G/499 Hence,

RA=GSG+S=G500{R_A} = {{GS} \over {G + S}} = {G \over {500}}
Q33
Two identical long conducting wires AOBAOB and CODCOD are placed at right angle to each other, with one above other such that OO is their common point for the two. The wires carry II 1 and II 2 currents, respectively. Point PP is lying at distance f from OO along a direction perpendicular to the plane containing the wires. The magnetic field at the point PP will be
A μ02πd(I1I2){{{\mu _0}} \over {2\pi d}}\left( {{{{I_1}} \over {{I_2}}}} \right)
B μ02πd(I1+I2){{{\mu _0}} \over {2\pi d}}\left( {{I_1} + {I_2}} \right)
C μ02πd(I12−I22){{{\mu _0}} \over {2\pi d}}\left( {I_1^2 - I_2^2} \right)
D μ02πd(I12+I22)1/2{{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{1/2}}
Correct Answer
Option D
Solution
B1=μ0I12πd(−j^){B_1} = {{{\mu _0}{I_1}} \over {2\pi d}}\left( { - \widehat j} \right)
B2=μ0I22πd(−i^){B_2} = {{{\mu _0}{I_2}} \over {2\pi d}}\left( { - \widehat i} \right)
B=B12+B22B = \sqrt {B_1^2 + B_2^2}
=μ02πd×I12+I22= {{{\mu _0}} \over {2\pi d}} \times \sqrt {I_1^2 + I_2^2}
Q34
A circular coil ABCD carrying a current 'i' is placed in a uniform magnetic field. If the magnetic force on the segment AB is F→\overrightarrow F , the force on the remaining segment BCDA is
A −F→ - \overrightarrow F
B 3F→3\overrightarrow F
C −-3F→3\overrightarrow F
D F→\overrightarrow F
Correct Answer
Option A
Solution

Here,

F→AB+F→BCDA=0→{\overrightarrow F _{AB}} + {\overrightarrow F _{BCDA}} = \overrightarrow 0
⇒F→BCDA=−F→AB=−F→\Rightarrow {\overrightarrow F _{BCDA}} = - {\overrightarrow F _{AB}} = - \overrightarrow F

(∵\because F AB =

F→\overrightarrow F

)

Q35
A long straight wire carries a certain current and produces a magnetic field 2 × \times 10 −-4 Wb m −-2 at a perpendicular distance of 5 cm from the wire. An electron situated at 5 cm from the wire moves with a velocity 10 7 m/s towards the wire along perpendicular to it. The force experienced by the electron will be (charge on electron 1.6 × \times 10 −-19 C)
A 3.2 N
B 3.2 × \times 10 −-16 N
C 1.6 × \times 10 −-16 N
D zero
Correct Answer
Option B
Solution

The situation is as shown in the figure.

Here, v = 10 7 m/s, B = 2 × 10 –4 Wb/m 2 The magnitude of the force experienced by the electron is F = evBsinθ\theta (

∵v→ and B→ are\because {\overrightarrow v \,and\,\overrightarrow B \,are}

perpendicular to each other) = evBsin90° = 1.6 × 10 –19 × 10 7 × 2 × 10 –4 × 1 = 3.2 × 10 –16 N

Q36
When a proton is released from rest in a room, it starts with an initial acceleration aa 0 towards west. When it is projected towards north with a speed vv 0 it moves with an initial acceleration 3aa 0 towards west. The an initial accelearation 3a 0 towards west. The an initial acceleration 3aa 0 toward west. The electric and magnetic fields in the room are
A ma0e{{m{a_0}} \over e} east, 3ma0ev0{{3m{a_0}} \over {e{v_0}}} up
B ma0e{{m{a_0}} \over e} east, 3ma0ev0{{3m{a_0}} \over {e{v_0}}} down
C ma0e{{m{a_0}} \over e} west, 2ma0ev0{{2m{a_0}} \over {e{v_0}}} up
D ma0e{{m{a_0}} \over e} west, 2ma0ev0{{2m{a_0}} \over {e{v_0}}} down
Correct Answer
Option D
Solution

When moves with an acceleration a 0 towards west, electric field

E=Fq=ma0e(West)E = {F \over q} = {{m{a_0}} \over e}\left( {West} \right)

When moves with an acceleration 3a 0 towards east, magnetic field

B=2ma0ev0(downward)B = {{2m{a_0}} \over {e{v_0}}}\left( {downward} \right)
Q37
A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α\alpha -particle to describe a circle of same radius in the same field?
A 2 MeV
B 1 MeV
C 0.5 MeV
D 4 MeV
Correct Answer
Option B
Solution

According to the principal of circular motion in a magnetic field

Fc=Fm⇒mv2R=qVB{F_c} = {F_m} \Rightarrow {{m{v^2}} \over R} = qVB
⇒R=mvqB=PqB=2m.kqB\Rightarrow R = {{mv} \over {qB}} = {P \over {qB}} = {{\sqrt {2m.k} } \over {qB}}
Rα=2(4m)K′2qB{R_\alpha } = {{\sqrt {2\left( {4m} \right)K'} } \over {2qB}}
RRα=KK′{R \over {{R_\alpha }}} = \sqrt {{K \over {K'}}}

but

R=RαR = {R_\alpha }

(given) ∴\therefore K = K' = 1 MeV

Q38
An alternating electric field, of frequency vv, is applied across the does (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by
A B=mυeB = {{m\upsilon } \over e} and K=2mπ2υ2R2K = 2m{\pi ^2}{\upsilon ^2}{R^2}
B B=2πmυeB = {{2\pi m\upsilon } \over e} K=m2πυR2K = {m^2}\pi \upsilon {R^2}
C B=2πmυeB = {{2\pi m\upsilon } \over e} K=2mπ2v2R2K = 2m{\pi ^2}{v^2}{R^2}
D B=mυeB = {{m\upsilon } \over e} K=m2πυR2K = {m^2}\pi \upsilon {R^2}
Correct Answer
Option C
Solution

Time period of cyclotron is

T=1v=2πmeB;B=2πmev;R=mveB=peBT = {1 \over v} = {{2\pi m} \over {eB}}; B = {{2\pi m} \over e}v;R = {{mv} \over {eB}} = {p \over {eB}}
⇒P=eBR=e×2πmveR=2πmvR\Rightarrow P = eBR = e \times {{2\pi mv} \over e}R = 2\pi mvR
K.E.=p22m=(2πmvR)22m=2π2mv2R2K.E. = {{{p^2}} \over {2m}} = {{{{\left( {2\pi mvR} \right)}^2}} \over {2m}} = 2{\pi ^2}m{v^2}{R^2}
Q39
Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are II and 2II, respectively. The resultant magnetic field induction at the centre will be
A 5μ0I2R{{\sqrt 5 {\mu _0}I} \over {2R}}
B 5μ0IR{{\sqrt 5 {\mu _0}I} \over R}
C μ0I2R{{{\mu _0}I} \over {2R}}
D μ0IR{{{\mu _0}I} \over R}
Correct Answer
Option A
Solution

Magnetic field induction due to vertical loop at the centre O is

B1=μ0I2R{B_1} = {{{\mu _0}I} \over {2R}}

It acts in horizontal direction. Magnetic field induction due to horizontal loop at the centre O is

B2=μ02I2R{B_2} = {{{\mu _0}2I} \over {2R}}

It acts in vertically upward direction.

As B 1 and B 2 are perpendicular to each other, therefore the resultant magnetic field induction at the centre O is

Bnet=B12+B22{B_{net}} = \sqrt {B_1^2 + B_2^2}
=(μ0I2R)2+(μ02I2R)2= \sqrt {{{\left( {{{{\mu _0}I} \over {2R}}} \right)}^2} + {{\left( {{{{\mu _0}2I} \over {2R}}} \right)}^2}}
Bnet=μ0I2R(1)2+(2)2=5μ0I2R{B_{net}} = {{{\mu _0}I} \over {2R}}\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} = {{\sqrt 5 {\mu _0}I} \over {2R}}
Q40
A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampare range. The value (in ohm) of neccessary shunt will be
A 0.001
B 0.01
C 1
D 0.05
Correct Answer
Option A
Solution
S=Vg(I−Ig)S = {{{V_g}} \over {\left( {I - {I_g}} \right)}}

Neglecting I g ∴\therefore

S=VgI=25×10−3V25 A=0.001ΩS = {{{V_g}} \over I} = {{25 \times {{10}^{ - 3}}V} \over {25\,A}} = 0.001\Omega
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