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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 385 questions
31 Medium
31 of 385
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.
32 Medium
32 of 385
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.
33 Medium
33 of 385
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.
34 Medium
34 of 385
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}.
35 Medium
35 of 385
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.
36 Medium
36 of 385
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.
37 Medium
37 of 385
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.
38 Medium
38 of 385
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.
39 Medium
39 of 385
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.
40 Medium
40 of 385
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.
41 Medium
41 of 385
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.
42 Medium
42 of 385
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.
43 Medium
43 of 385
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.
44 Medium
44 of 385
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.
45 Medium
45 of 385
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.
46 Medium
46 of 385
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.
47 Medium
47 of 385
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.
48 Medium
48 of 385
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.
49 Medium
49 of 385
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.
50 Medium
50 of 385
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.
51 Medium
51 of 385
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.
52 Medium
52 of 385
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.
53 Medium
53 of 385
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.
54 Medium
54 of 385
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.
55 Medium
55 of 385
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.
56 Medium
56 of 385
Understanding +4 | -1
A charge moving with velocity v in x-direction
is subjected to a field of magnetic induction in
negative x-direction. As a result, the charge will
View answer and explanation
Correct option: A
remain unaffected
Option A (remain unaffected) is correct because the magnetic force on a moving charge is given by F→=q(v→×B→)\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}), and when the velocity and magnetic field are parallel or anti-parallel, the force is zero.
57 Medium
57 of 385
Application +4 | -1
A thin circular disc of radius RR is uniformly charged with density σ\sigma. It rotates about its diameter with angular speed ω\omega. The magnetic moment of the disc is
View answer and explanation
Correct option: D
πR48σω\frac{\pi R^4}{8} \sigma \omega
Option D (πR48σω\frac{\pi R^4}{8} \sigma \omega) is correct because it is derived from the relationship between magnetic moment, angular momentum, charge, and mass.
58 Medium
58 of 385
Understanding +4 | -1
The angle made by orbital angular momentum of electron with the direction of the orbital magnetic moment is
View answer and explanation
Correct option: B
180∘180^{\circ}
Option B (If both assertion and reason are true but reason is not correct explanation of the assertion) is correct because both statements are true, but the reason (polarization) doesn't directly explain why sound cannot travel in a vacuum.
59 Medium
59 of 385
Understanding +4 | -1
The direction of magnetic field at a point due
to an infinitely long wire carrying current is :
View answer and explanation
Correct option: D
perpendicular to the plane containing the
conductor and the point
Option D is correct because the magnetic field due to a current-carrying wire is perpendicular to both the wire and the line connecting the point to the wire, according to the right-hand rule.
60 Medium
60 of 385
Application +4 | -1
Two long and parallel straight wires AA and BB carrying current 5 A5 \mathrm{~A} each in same direction. If the force on a 5 cm5 \mathrm{~cm} section of wire BB is 5×10−65 \times 10^{-6} N\mathrm{N}, then the separation between two wires is ( μ0=4π×10−7\mu_0=4 \pi \times 10^{-7} SI unit)
View answer and explanation
Correct option: A
2.5 cm2.5 \mathrm{~cm}
Option A (2.5 cm) is correct because it correctly applies the formula for the force between two parallel wires.
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