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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Mixed difficulty

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Showing 31-60 of 418 questions

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31 Medium
31 of 418
Application +4 | -1
An electron of mass mm is accelerated by a d.c. potential difference VV and then subjected to a transverse magnetic field BB, The trajectory of the electron will be a circle of radius
View answer and explanation
Correct option: D
2mVeB2\sqrt{\frac{2 m V}{eB^2}}
Option D (2mVeB2\sqrt{\frac{{2 m V}}{{eB^2}}}) is correct because it is derived from equating the kinetic energy gained by the electron to its potential energy and then using the formula for the radius of a charged particle in a magnetic field.
32 Medium
32 of 418
Understanding +4 | -1
Assertion: Free electron always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it.

Reason: The average velocity of free electron is zero.
View answer and explanation
Correct option: A
(A) and (R) are true and (R) is the correct explanation of (A)
Option A ((A) and (R) are true and (R) is the correct explanation of (A)) is correct because free electrons are always in motion, but their average velocity is zero, resulting in no net magnetic force unless a current is present.
33 Medium
33 of 418
Application +4 | -1
A wire of length ' LL ' carries current ' II ' along xx axis, A magnetic field B=B0(i^−j^−k^)TB=B_0(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \mathbf{T} acts on the wire. The magnitude of magnetic force acting on the wire is
View answer and explanation
Correct option: C
2ILB0\sqrt{2} \mathrm{ILB}_0
Option C (2ILB0\sqrt{2} \mathrm{ILB}_0) is correct because it's the magnitude of the cross product between the current element and the magnetic field.
34 Medium
34 of 418
Understanding +4 | -1
Assertion: The energy of charged particle moving in a uniform magnetic field does not change.

Reason: Work done by magnetic field on the charge is zero.
View answer and explanation
Correct option: A
(A) and (R) are true and (R) is the correct explanation of (A)
Option A ((A) and (R) are true and (R) is the correct explanation of (A)) is correct because the magnetic force is always perpendicular to the velocity, resulting in zero work done and thus no change in energy.
35 Medium
35 of 418
Understanding +4 | -1
When a charged particle enters uniform
magnetic field its kinetic energy
View answer and explanation
Correct option: A
remains constant
Option A (remains constant) is correct because the magnetic force on a moving charged particle is always perpendicular to its velocity, doing no work and thus not changing its kinetic energy.
36 Medium
36 of 418
Application +4 | -1
A straight wire carries a current of 3 A3 \mathrm{~A}. Calculate the magnitude of the magnetic field at a point 10 cm10 \mathrm{~cm} away from the wire.
View answer and explanation
Correct option: A
6×10−6 T6 \times 10^{-6} \mathrm{~T}
Option C (01:08) is correct because the time period is proportional to n cubed, where n is the orbit number.
37 Medium
37 of 418
Understanding +4 | -1
A charged particle enters a uniform magnetic
field with a certain speed at right angles to it.
In the magnetic field a change could occur in
its
View answer and explanation
Correct option: C
linear momentum
Option C (linear momentum) is correct because the magnetic force changes the direction of the velocity, thus changing the linear momentum, but not the speed or kinetic energy.
38 Medium
38 of 418
Application +4 | -1
A very long conductor carries a current of 1 A. The magnetic field at a point whose distance is 0.1 m0.1 \mathrm{~m} from one end of the conductor is (in ' TT ')
View answer and explanation
Correct option: A
10−610^{-6}
Option A (10−610^{-6}) is correct because the magnetic field is calculated using the formula B=μ0I4πrB=\frac{\mu_0 I}{4 \pi r}.
39 Medium
39 of 418
Application +4 | -1
A solenoid is 2.0 m2.0 \mathrm{~m} long and 3.0 cm3.0 \mathrm{~cm} in diameter. It has 5 layers of winding of 1000 turns each and carries a current of 5.0 A5.0 \mathrm{~A}. What is the magnetic field at the centre? Use the standard value of μ0\mu_0.
View answer and explanation
Correct option: A
1.57×10−2 T1.57 \times 10^{-2} \mathrm{~T}
Option A (1.57×10−2 T1.57 \times 10^{-2} \mathrm{~T}) is correct because it is the magnetic field calculated using the formula B=μ0nIB = \mu_0 n I with the given parameters.
40 Medium
40 of 418
Analyzing +4 | -1
A current carrying loop lies on a smooth horizontal plane. Then,
View answer and explanation
Correct option: B
It is possible to establish a uniform magnetic field in the region so that the loop will tip over about any of the point.
Option B (It is possible to establish a uniform magnetic field in the region so that the loop will tip over about any of the point) is correct because a torque can be generated to cause tipping.
41 Medium
41 of 418
Understanding +4 | -1
Assertion (A): The range of given voltmeter can be both increased and decreased.

Reason (R): By adjusting the value of resistance in series with galvanometer the range of voltmeter can be adjusted.
View answer and explanation
Correct option: A
(A) and (R) are true and (R) is the correct explanation of (A)
Option A ((A) and (R) are true and (R) is the correct explanation of (A)) is correct because the range of a voltmeter can be adjusted by changing the series resistance, and the given reason accurately explains this.
42 Medium
42 of 418
Application +4 | -1
A negatively charged particle projected
towards east is deflected towards north by a
magnetic filed. The field may be:
View answer and explanation
Correct option: D
Downwards
Option D (Downwards) is correct because, according to Fleming's left-hand rule, a negatively charged particle moving east and deflected north requires a downward magnetic field.
43 Medium
43 of 418
Remembering +4 | -1
The formula for magnetic field of a certain component is given by B=μ0Ni2πrB=\frac{\mu_0 N i}{2 \pi r}

Where, N=\mathbf{N}= total number of turns
i=i= current through it
r={r}= radius.

The component is
View answer and explanation
Correct option: C
torrid
The correct answer is torrid because the magnetic field formula B=μ0Ni2πrB=\frac{\mu_0 N i}{2 \pi r} is the formula for the magnetic field inside a toroid.
44 Medium
44 of 418
Understanding +4 | -1
An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?
View answer and explanation
Correct option: D
The electron will continue to move with uniform velocity along the axis of the solenoid
Option D (The electron will continue to move with uniform velocity along the axis of the solenoid) is correct because the magnetic force on the electron is zero since the velocity is parallel to the magnetic field.
45 Medium
45 of 418
Application +4 | -1
An electron having kinetic energy 10eV10 \mathrm{eV} is circulating in a path of radius 0.1 m0.1 \mathrm{~m} in an external magnetic field of intensity 10−4 T10^{-4} \mathrm{~T}. The speed of the electron will be
View answer and explanation
Correct option: A
1.76×(106)ms−11.76 \times\left(10^6\right) \mathrm{ms}^{-1}
Option A (1.76×(106)ms−11.76 \times(10^6) \mathrm{ms}^{-1}) is correct because it is the speed calculated using the formula relating kinetic energy, magnetic field, radius, and charge of the electron.
46 Medium
46 of 418
Understanding +4 | -1
Two beams of electrons are moving parallel to each other then magnetic force acting between them is
View answer and explanation
Correct option: A
Attractive
Option A is correct because parallel electron beams moving in the same direction constitute parallel currents, which attract each other magnetically.
47 Medium
47 of 418
Application +4 | -1
A circular coil of diameter 7 cm7 {~cm} has 24 turns of wire carrying current of 0.75 A0.75 {~A}. The magnetic moment of the coil is
View answer and explanation
Correct option: A
6.9×10−2amp−m26.9 \times 10^{-2} a m p-m^2
Option A (6.9×10−2amp−m26.9 \times 10^{-2} a m p-m^2) is correct because it is the result of calculating the magnetic moment using the formula M = NiA.
48 Medium
48 of 418
Remembering +4 | -1
The force on a conductor of length ll placed in a magnetic field of magnitude BB and carrying a current ii is given by ( θ\theta is the angle, the conductor makes with the direction of BB ) :
View answer and explanation
Correct option: A
F=il Bsin⁡θ\mathrm{F}=\mathrm{i} l \mathrm{~B} \sin \theta
Option A (F=il Bsin⁡θ\mathrm{F}=\mathrm{i} l \mathrm{~B} \sin \theta) is correct because it represents the magnitude of the force on a current-carrying conductor in a magnetic field, where θ\theta is the angle between the conductor and the magnetic field.
49 Medium
49 of 418
Application +4 | -1
A proton and an α−\alpha \mathrm{ - } particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field B is set up perpendicular to their velocities, the ratio of the radii rp:rα{r_p}:{r_\alpha } of the circular paths described by them will be
View answer and explanation
Correct option: A
1:21:\sqrt 2
Option A (1:21:\sqrt 2) is correct because the radius of the circular path is proportional to m/q\sqrt{m/q}, where m is mass and q is charge.
50 Hard
50 of 418
Application +4 | -1
A constant voltage is applied to a series R−L\mathrm{R}-\mathrm{L} circuit by closing the switch. The voltage across inductor (L=2H)(\mathrm{L}=2 \mathrm{H}) is 20 V20 \mathrm{~V} at t=0\mathrm{t}=0 and drops to 5 V5 \mathrm{~V} at 20 ms20 \mathrm{~ms}. The value of R\mathrm{R} in Ω\Omega is:
View answer and explanation
Correct option: C
100ln⁡4Ω100 \ln 4 \Omega
Option C (100ln⁡4Ω100 \ln 4 \Omega) is correct because it is the value of R calculated using the given voltage drop across the inductor in an RL circuit.
51 Medium
51 of 418
Understanding +4 | -1
A charged particle moves through a magnetic
field in a direction perpendicular to it. Then the
View answer and explanation
Correct option: B
speed of the particle remains unchanged
The correct answer is speed of the particle remains unchanged because the magnetic force is always perpendicular to the velocity, doing no work and thus not changing the speed.
52 Medium
52 of 418
Understanding +4 | -1
The frequency of the charged particle circular at right angles to a uniform magnetic field does not depend upon the
View answer and explanation
Correct option: A
Speed of the particle
Option A (Speed of the particle) is correct because the frequency of a charged particle in a uniform magnetic field is independent of its speed.
53 Medium
53 of 418
Understanding +4 | -1
In a chamber, a uniform magnetic field of 6.5 G(1 G=6.5\,\mathrm{G}(1\,\mathrm{G}= 10−4 T10^{-4} \mathrm{~T} ) is maintained. An electron is shot into the field with a speed of 4.8×106 ms−14.8 \times 10^6 \mathrm{~ms}^{-1} normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. Given that e=1.6×10−19C,me=9.1×10−31 kge=1.6 \times 10^{-19} \mathrm{C},\quad m_e=9.1 \times 10^{-31} \mathrm{~kg}.
View answer and explanation
Correct option: A
4.2 cm
Option A (4.2 cm) is correct because the radius of the circular path is calculated using the formula r=meveBr = \frac{{m_e v}}{{e B}} with the given values.
54 Medium
54 of 418
Understanding +4 | -1
Biot-Savart law indicates that the moving electrons (velocity v{v} ) produce a magnetic field B{B} such that
View answer and explanation
Correct option: A
B⊥v{B} \perp {v}
Option A (B⊥v{B} \perp {v}) is correct because the magnetic field produced by a moving charge is perpendicular to both the velocity of the charge and the position vector from the charge to the point of observation, as described by the Biot-Savart law.
55 Medium
55 of 418
Understanding +4 | -1
In order to increase the sensitivity of galvanometer
View answer and explanation
Correct option: C
The magnetic field should be increased
Option C (The magnetic field should be increased) is correct because the sensitivity of a galvanometer is directly proportional to the magnetic field strength.
56 Medium
56 of 418
Application +4 | -1
A galvanometer coil has a resistance of 10Ω10 \Omega and the ammeter shows full scale deflection for a current of 1 mA1 {~mA}. The shunt resistance required to convert the galvanometer into an ammeter of range 0.100 mA0.100 {~mA} is about
View answer and explanation
Correct option: C
0.1 Ω0.1 \,\Omega
Option C (0.1 Ω0.1 \,\Omega) is correct because it is calculated using the formula for shunt resistance, ensuring the galvanometer can measure the desired higher current range.
57 Medium
57 of 418
Remembering +4 | -1
The intensity of magnetic field due to current I in a long straight wire is proportional to
View answer and explanation
Correct option: A
I
Option A (I) is correct because the magnetic field intensity due to a long straight wire is directly proportional to the current flowing through the wire.
58 Medium
58 of 418
Application +4 | -1
Two particles having same charge and KE enter at right angles into the same magnetic field and travel in circular paths of radii 2 cm and 3 cm respectively. The ratio of their masses is
View answer and explanation
Correct option: A
0.172916667
Option A (04:09) is correct because the ratio of masses is proportional to the square of the radii when charge and KE are the same.
59 Medium
59 of 418
Application +4 | -1
A proton and an alpha particle both enter a region of uniform magnetic field B\mathrm{B}, moving at right angles to the field B\mathrm{B}. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV\mathrm{1}\,\mathrm{MeV}, the energy acquired by the alpha particle will be:
View answer and explanation
Correct option: A
1
Option A (1) is correct because the kinetic energy of the alpha particle is the same as the proton when they have the same radius in the same magnetic field.
60 Medium
60 of 418
Understanding +4 | -1
When a current carrying conductor is placed in magnetic field,then:
View answer and explanation
Correct option: B
It may experience a force
Option B (It may experience a force) is correct because the force on a current-carrying conductor in a magnetic field depends on the angle between the current and the magnetic field.
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