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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

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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 331-360 of 385 questions

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331 Medium
331 of 385
Analyzing +4 | -1
Assertion: Various regular polygons formed by a constant length current carrying wire, then magnetic induction at their centres are increases when their number of sides increases.
Reason: A magnetic dipole placed in non uniform magnetic field, then torque acting on it is always zero.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because both the assertion and the reason are false statements regarding magnetic induction and torque on a magnetic dipole.
332 Medium
332 of 385
Understanding +4 | -1
Magnetic moment for a solenoid and corresponding bar magnet is
View answer and explanation
Correct option: A
equal for both
Option A (equal for both) is correct because a solenoid and a corresponding bar magnet produce the same magnetic field, resulting in equal magnetic moments.
333 Medium
333 of 385
Application +4 | -1
A galvanometer has a resistance of 96 Ω96\,\Omega and it is desired to pass 4%4 \% of the total current through it. Calculate the value of shunt resistance.
View answer and explanation
Correct option: A
4 Ω4\,\Omega
Option A (4 Ω4\,\Omega) is correct because it is the shunt resistance value calculated using the current division rule for parallel resistors.
334 Medium
334 of 385
Application +4 | -1
A charge particle of charge q is moving in a circle of radius R with uniform speed v. Magnetic moment M is given by
View answer and explanation
Correct option: B
qvR2\frac{{qvR}}{2}
Option B (qvR2\frac{{qvR}}{2}) is correct because the magnetic moment is directly proportional to the charge, velocity, and radius, and inversely proportional to 2.
335 Medium
335 of 385
Application +4 | -1
A straight horizontal wire of mass 10mg10 \mathrm{mg} and length 1.0 m1.0 \mathrm{~m} carries a current of 2.0 A2.0 \mathrm{~A}. What minimum magnetic field BB should be applied in the region so that the magnetic force on the wire may balance its weight?
View answer and explanation
Correct option: D
4.9×10−5 T4.9 \times 10^{-5} \mathrm{~T}
Option D (4.9×10−5 T4.9 \times 10^{-5} \mathrm{~T}) is correct because it is the minimum magnetic field required to balance the weight of the wire, calculated using Mg=ilBMg = ilB.
336 Medium
336 of 385
Understanding +4 | -1
A charged particle experiences magnetic force
in the presence of magnetic field. Which of the
following statement is correct ?
View answer and explanation
Correct option: A
The particle is moving and magnetic field is
perpendicular to the velocity
The correct answer is A (The particle is moving and magnetic field is perpendicular to the velocity) because the magnetic force is proportional to q(v×B)q(v \times B).
337 Medium
337 of 385
Understanding +4 | -1
A current-carrying wire is known to produce
magnetic lines of force around the conducting
straight wire. The direction of the lines of force
may be described by:
View answer and explanation
Correct option: C
right-hand thumb rule for both up and down
currents
Option C (right-hand thumb rule for both up and down currents) is correct because the right-hand thumb rule describes the direction of magnetic field lines around a current-carrying wire, regardless of the current's direction.
338 Medium
338 of 385
Remembering +4 | -1
A charge q moves with velocity ′V⃗′{ }^{\prime} \vec{V}{ }^{\prime} through electric field (E→)(\overrightarrow{\mathbf{E}}) as well as magnetic field (B→)(\overrightarrow{\mathbf{B}}). Then the force acting on it is
View answer and explanation
Correct option: B
qE→+q(V→×B→)\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})
Option B (qE→+q(V→×B→)\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})) is correct because it represents the Lorentz force, which is the sum of the electric and magnetic forces on a moving charge.
339 Medium
339 of 385
Remembering +4 | -1
Tangent galvanometer measures
View answer and explanation
Correct option: B
Current
Option B (Current) is correct because a tangent galvanometer is specifically designed to measure electric current based on the tangent law.
340 Medium
340 of 385
Remembering +4 | -1
The magnetic field at the center of current I carrying loop of radius rr, is
View answer and explanation
Correct option: A
μ0I2r\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}
Option A (μ0I2r\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}) is correct because the magnetic field at the center of a current loop is given by this formula.
341 Medium
341 of 385
Understanding +4 | -1
A straight current carrying conductor is kept
along the axis of circular loop carrying current.
The force exerted by the straight conductor on
the loop is :
View answer and explanation
Correct option: A
zero
Option A is correct because the magnetic field due to the straight current-carrying conductor is parallel to the plane of the loop, resulting in zero force on the loop.
342 Medium
342 of 385
Application +4 | -1
Into a transverse uniform magnetic field of induction 6.5 G an electron is projected with a speed 4.8×106m/s4.8\mathrm{ } \times \mathrm{ }{10^6}m/s, at angle 90∘90^\circ to the boundary. The time elapsed by the electron in the field in nano second is
View answer and explanation
Correct option: C
28.1
Option C (28.1) is correct because the time elapsed is half the period of the circular motion of the electron in the magnetic field.
343 Medium
343 of 385
Understanding +4 | -1
If number of turns in moving coil galvanometer
becomes half, then the deflection for the same
current will become
View answer and explanation
Correct option: D
half
Option D (half) is correct because the deflection in a moving coil galvanometer is directly proportional to the number of turns.
344 Medium
344 of 385
Understanding +4 | -1
A galvanometer acting as a voltmeter should have
View answer and explanation
Correct option: C
High resistance in series in series with its coil
Option C (High resistance in series in series with its coil) is correct because a voltmeter needs high resistance to minimize current drawn from the circuit being measured, achieved by adding a high resistance in series with the galvanometer coil.
345 Medium
345 of 385
Application +4 | -1
A voltmeter has a resistance of G  ohmG\,\,ohm and range VV volt. The value of resistance used in series to convert it into a voltmeter of range nVnV volt is
View answer and explanation
Correct option: C
(n−1)G\left( {n - 1} \right)G
Option C ((n-1)G) is correct because the series resistance required to increase the range of a voltmeter is (n-1)G, where G is the original resistance and n is the factor by which the range is increased.
346 Medium
346 of 385
Remembering +4 | -1
A charge q{q} moves in a region where electric field and magnetic field both exist, then force on it is
View answer and explanation
Correct option: D
qE⃗+q(v⃗×B⃗)q \vec{E}+q(\vec{v} \times \vec{B})
Option D (qE⃗+q(v⃗×B⃗)q \vec{E}+q(\vec{v} \times \vec{B})) is correct because it represents the Lorentz force, which is the sum of the electric and magnetic forces on a moving charge.
347 Medium
347 of 385
Understanding +4 | -1
The magnitude of the magnetic field produced by a current carrying loop at a large distance xx varies as
View answer and explanation
Correct option: C
1x3\frac{1}{\mathrm{x}^3}
Option C (1x3\frac{1}{\mathrm{x}^3}) is correct because the magnetic field due to a current loop at a large distance varies inversely with the cube of the distance.
348 Medium
348 of 385
Analyzing +4 | -1
Statement A: When a proton and a neutron enter into a transverse magnetic field with equal speeds, then they trace circular paths of equal radii.
Statement B: In a transverse magnetic field the period of revolution of a charged particle in a circular path is directly proportional to the mass of the particle.
View answer and explanation
Correct option: D
A is not correct and B is correct
Statement A is incorrect because neutrons are uncharged and do not experience a force in a magnetic field. Statement B is correct because the period of revolution is directly proportional to mass.
349 Medium
349 of 385
Understanding +4 | -1
A charged particle moves through a magnetic
field in a direction perpendicular to it. Then it
View answer and explanation
Correct option: B
speed of the particle remains unchanged
Option B is correct because when a charged particle moves perpendicular to a magnetic field, its speed remains unchanged, only the direction changes.
350 Medium
350 of 385
Application +4 | -1
A portion is rotating along a circular path with kinetic energy KK in a uniform magnetic field BB. If the magnetic field is made four times, the kinetic energy of rotation of proton is
View answer and explanation
Correct option: D
K
Option D (K) is correct because the kinetic energy of the proton remains constant as the magnetic field changes, given the radius adjusts accordingly.
351 Medium
351 of 385
Understanding +4 | -1
The force on a charged particle moving with a velocity v→\overrightarrow{\mathbf{v}} in a magnetic field B⃗\vec{B} is :
View answer and explanation
Correct option: B
maximum, if v→\overrightarrow{\mathrm{v}} perpendicular to B→\overrightarrow{\mathrm{B}}
Option B (maximum, if v→\overrightarrow{\mathrm{v}} perpendicular to B→\overrightarrow{\mathrm{B}}) is correct because the magnetic force is proportional to the sine of the angle between velocity and magnetic field, which is maximum at 90 degrees.
352 Medium
352 of 385
Understanding +4 | -1
A charge QQ is moving dldl distance in the magnetic field BB.find the value of work done by BB.
View answer and explanation
Correct option: C
Zero
Option C is correct because the work done by a magnetic field on a moving charge is always zero, as the magnetic force is perpendicular to the direction of motion.
353 Medium
353 of 385
Application +4 | -1
A galvanometer, having a resistance of 50  Ω\mathrm{50}\,\,\mathrm{\Omega }, gives a full scale deflection for a current of 0.05  A0.05\,\,A. The length in meter of a resistance wire of area of cross-section 2.97  ×  10−2  cm22.97\,\, \times \,\,{10^{ - 2}}\,\,\mathrm{c}{\mathrm{m}^2} that can be used to convert the galvanometer into a ammeter which can read a maximum of 5  A\mathrm{5}\,\,\mathrm{A} current is (Specific resistance of the wire =5  ×  10−7  Ω m= 5\,\, \times \,\,{10^{ - 7}}\,\,\mathrm{\Omega }\,\mathrm{m}
View answer and explanation
Correct option: C
3
Option C (3) is correct because the length of the resistance wire is calculated using the shunt resistance formula and the specific resistance of the wire.
354 Medium
354 of 385
Analyzing +4 | -1
In case of a charged particle moving in a non-uniform magnetic field (independent of time)
View answer and explanation
Correct option: D
Momentum changes
Option D (Momentum changes) is correct because in a non-uniform magnetic field, the direction of the force on the charged particle changes, leading to a change in momentum, even if speed remains constant.
355 Medium
355 of 385
Application +4 | -1
A charged particle is accelerated through potential V0V_0 is subjected to a crossed electric field (E)(E) and magnetic field (B)(B). The field are adjusted such that the charged particle is not deflected. The specific charge of particle is given by
View answer and explanation
Correct option: A
E22B2V0\frac{E^2}{2 B^2 V_0}
Option A (E22B2V0\frac{{E^2}}{{2 B^2 V_0}}) is correct because it represents the specific charge derived from equating electric and magnetic forces and using the kinetic energy gained from the potential difference.
356 Medium
356 of 385
Understanding +4 | -1
An electron moving with constant velocity passes through a region of space without change in its velocity. If EE and BB represent electric and magnetic field respectively, the region of space may have
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the electron can maintain constant velocity if there are no fields, if it moves parallel to the magnetic field, or if electric and magnetic forces cancel each other.
357 Medium
357 of 385
Application +4 | -1
A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20  Ω20\,\,\Omega. Calculate the resistance to be connected in series to convert it into a volt meter of resistance range.
View answer and explanation
Correct option: A
9980
Option A (9980) is correct because the series resistance needed to convert the galvanometer into a voltmeter is calculated using the formula R = G(n-1).
358 Medium
358 of 385
Understanding +4 | -1
A magnet is suspended lengthwise from a spring and while it oscillates, the magnet moves in and out of the coil C connected to a galvanometer G. Then as the magnet oscillates.
View answer and explanation
Correct option: D
Deflection of G to the left and right has decreasing amplitude
Option D is correct because as the magnet oscillates in and out of the coil, it induces a current in the coil, which causes the galvanometer to deflect left and right. The amplitude of the deflection decreases as the magnet's oscillations dampen due to energy loss.
359 Medium
359 of 385
Application +4 | -1
A current of 1 A1 \mathrm{~A} is flowing through a straight conductor of length 16 cm16 \mathrm{~cm}. The magnetic induction (in tesla) at a point 10 cm10 \mathrm{~cm} from the either end of the wire is:
View answer and explanation
Correct option: A
83×10−6\frac{8}{3} \times 10^{-6}
Option A (83×10−6\frac{8}{3} \times 10^{-6}) is correct because it is calculated using the formula for the magnetic field due to a straight conductor at a given distance.
360 Medium
360 of 385
Understanding +4 | -1
For the magnetic field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be
View answer and explanation
Correct option: B
90∘90^{\circ}
Option B is correct because the magnetic field due to a current element is maximum when the angle between the current element and the position vector is 90 degrees.
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