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Magnetic Effect of Current

NEET Physics · 96 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be
A 900 Ω\Omega
B 1800 Ω\Omega
C 500 Ω\Omega
D 1000 Ω\Omega
Correct Answer
Option A
Solution

Here, Resistance of galvanometer, G = 100

Ω\Omega

Current for full scale deflection, I g = 30 mA = 30 × 10 –3 A Range of voltmeter, V = 30 V To convert the galvanometer into an voltmeter of a given range, a resistance R is connected in series with it as shown in the figure.

From figure, V = I g (G + R)

⇒R=VIg−G=3030×10−3−100Ω\Rightarrow R = {V \over {{I_g}}} - G = {{30} \over {30 \times {{10}^{ - 3}}}} - 100\Omega

= 1000 – 100 = 900

Ω\Omega
Q52
Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency ff Hz. The magnitude of magnetic induction at the center of the ring is
A μ0qf2πR{{{\mu _0}qf} \over {2\pi R}}
B μ0qf2R{{{\mu _0}qf} \over {2R}}
C μ0q2fR{{{\mu _0}q} \over {2fR}}
D μ0q2πfR{{{\mu _0}q} \over {2\pi fR}}
Correct Answer
Option B
Solution

When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is I =

qT{q \over T}

= qf Magnetic field at the centre of the ring is B =

μ0I2R{{{\mu _0}I} \over {2R}}

=

μ0qf2R{{{\mu _0}qf} \over {2R}}
Q53
Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle
A depends on R and not on v
B is independent of both v and R
C depends on both v and R
D depends on v and not on R
Correct Answer
Option B
Solution

For the circular motion in a cyclotron,

qvB=mv2r⇒qB=mω=m×2πTqvB = {{m{v^2}} \over r} \Rightarrow qB = m\omega = {{m \times 2\pi } \over T}
∴T=2πmqB\therefore T = {{2\pi m} \over {qB}}

is independent of v and r.

Q54
The magnetic force acting on a charged particle of charge −-2 μ\mu C in a magnetic frield of 2 T acting in y direction, when the particle velocity is (2i^+3j^)×106 ms−1\left( {2\widehat i + 3\widehat j} \right) \times {10^6}\,m{s^{ - 1}}
A 4 N in z direction
B 8 N in y direction
C 8 N in z direction
D 8 N in −- z direction
Correct Answer
Option D
Solution

The magnetic force acting on the charged paraticle is given by

F→=q(v→×B→)\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)
=(−2×10−6)[{(2i^+3j^)×106}×(2j^)]= \left( { - 2 \times {{10}^{ - 6}}} \right)\left[ {\left\{ {\left( {2\widehat i + 3\widehat j} \right) \times {{10}^6}} \right\} \times \left( {2\widehat j} \right)} \right]
=−4(2k^)= - 4\left( {2\widehat k} \right)
=−8k^= - 8\widehat k

∴\therefore Force is of 8N along – z-axis.

Q55
A galvanometer havings a coil resistance of 60 Ω\Omega shows full scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents upto 5.0 amp by
A putting in series a resistance of 15 Ω\Omega
B putting in series a resistance of 240 Ω\Omega
C putting in parallel a resistance of 15 Ω\Omega
D putting in parallel a resistance of 240 Ω\Omega
Correct Answer
Option C
Solution

G = 60

Ω\Omega

, I g = 1.0A, I = 5A. Let S be the shunt resistance connected in parallel to galvanometer I g G = (I – I g ) S,

S=IgGI−Ig=15−1×60=1.0×604=15ΩS = {{{I_g}G} \over {I - {I_g}}} = {1 \over {5 - 1}} \times 60 = {{1.0 \times 60} \over 4} = 15\Omega

Thus by putting 15

Ω\Omega

in parallel, the galvanometer can be converted into an ammeter.

Q56
A galvanometer of resistance 50 Ω\Omega is connected to a battery of 3 V along with a resistance of 2950 Ω\Omega in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
A 60506050 Ω\Omega
B 4450 Ω4450\,\Omega
C 5050 Ω5050\,\Omega
D 5550 Ω5550\,\Omega
Correct Answer
Option B
Solution

Total initial resistance = R G + R 1 = (50 + 2950)

Ω\Omega

= 3000

Ω\Omega
ε=3V\varepsilon = 3V

∴\therefore Current =

3V3000Ω=1×10−3mA{{3V} \over {3000\Omega }} = 1 \times {10^{ - 3}}mA

If the deflection has to be reduced to 20 divisions, current i = 1 mA ×\times

23{2 \over 3}

as the full deflection scale for 1 mA = 30 divisions.

3V=3000Ω×1mA=xΩ×23mA3V = 3000\Omega \times 1mA = x\Omega \times {2 \over 3}mA
⇒x=3000×1×32=4500Ω\Rightarrow x = 3000 \times 1 \times {3 \over 2} = 4500\Omega

But the galvanometer resistance = 50

Ω\Omega

Therefore the resistance to be added = (4500 – 50)

Ω\Omega

= 4450

Ω\Omega

.

Q57
A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are F 1 , F 2 and F 3 respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is
A (F3−F1)2−F22\sqrt {{{\left( {{F_3} - {F_1}} \right)}^2} - F_2^2}
B F3−F1+F2{F_3} - {F_1} + {F_2}
C F3−F1−F2{F_3} - {F_1} - {F_2}
D (F3−F1)2+F22\sqrt {{{\left( {{F_3} - {F_1}} \right)}^2} + F_2^2}
Correct Answer
Option D
Solution

According to the figure the magnitude of force on the segment QM is F 3 – F 1 and PM is F 2 .

Therefore, the magnitude of the force on segment PQ is

(F3−F1)2+F22\sqrt {{{\left( {{F_3} - {F_1}} \right)}^2} + F_2^2}
Q58
A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of induction B→\overrightarrow B . After 3 seconds the kinetic energy of the particle will be
A T
B 4 T
C 3 T
D 2 T
Correct Answer
Option A
Solution

When a charged particle having a given K.E., T enters in a field of magnetic induction, which is perpendicular to its velocity, it takes a circular trajectory.

It does not increase in energy, therefore T is the K.E.

Q59
The resistance of an ammeter is 13 Ω\Omega and its scale is graduated for a current upto 100 amps. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 amperes by this meter. The value of shunt-resistance is
A 2 Ω\Omega
B 0.2 Ω\Omega
C 2 kΩ\Omega
D 20 Ω\Omega
Correct Answer
Option A
Solution

We know

IIS=1+GS{I \over {{I_S}}} = 1 + {G \over S}
750100=1+13S{{750} \over {100}} = 1 + {{13} \over S}
⇒S=2Ω\Rightarrow S = 2\Omega
Q60
Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle
A depends on R and not on v
B is independent of both v and R
C depends on both v and R
D depends on v and not on R
Correct Answer
Option B
Solution

For the circular motion in a cyclotron,

qvB=mv2r⇒qB=mω=m×2πTqvB = {{m{v^2}} \over r} \Rightarrow qB = m\omega = {{m \times 2\pi } \over T}
∴T=2πmqB\therefore T = {{2\pi m} \over {qB}}

is independent of v and r.

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