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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Hard

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1 Hard
1 of 14
Application +4 | -1
Two long parallel wires carry currents i1i_1 and i2i_2 such that i1>i2.i_1>i_2. When the currents are in the same direction, the magnetic field at a point midway between the wires is 6×10−6 T.6 \times 10^{-6} \mathrm{~T}. If the direction of i2i_2 is reversed, the field becomes 3×10−5 T.3 \times 10^{-5} \mathrm{~T}. The ratio i1i2\frac{i_1}{i_2} is
View answer and explanation
Correct option: D
32\frac{3}{2}
The correct answer is D (32\frac{3}{2}) because it's derived from setting up two equations for the magnetic field, one for currents in the same direction and another for opposite directions, then solving for the ratio.
2 Hard
2 of 14
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.
3 Hard
3 of 14
Understanding +4 | -1
Assertion: The kinetic energy does not change when a particle moves in uniform magnetic field.
Reason: The speed of the particle is not affected by magnetic field.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A (If both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because the magnetic force is perpendicular to the velocity, so it does no work, keeping kinetic energy constant.
4 Hard
4 of 14
Application +4 | -1
A solenoid of 1000 turns per meter has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is (permeability of free space =  4π×10−7H/m= \,\,4\pi \times {10^{ - 7}}\mathrm{H/m} )
View answer and explanation
Correct option: A
πT\pi T
Option A (πT\pi T) is correct. The magnetic flux density is calculated using the formula B = μ_r μ_0 n I, where μ_r is the relative permeability, μ_0 is the permeability of free space, n is the number of turns per meter, and I is the current.
5 Hard
5 of 14
Application +4 | -1
The magnetic field due to current carrying a circular loop of radius 5 cm5 \mathrm{~cm} at a point on the axis at a distance of 12 cm12 \mathrm{~cm} from the center is 250μT250 \mu \mathrm{T}. The magnetic field at the center of the loop is
View answer and explanation
Correct option: B
4394μT4394 \mu \mathrm{T}
Option B (4394 μ T) is correct because it is the calculated magnetic field at the center of the loop, based on the given magnetic field at a point on the axis.
6 Hard
6 of 14
Analyzing +4 | -1
A charged particle of unit mass and unit charge moves with velocity of v⃗=(8i^+6j^)m/s\vec{v}=(8 \hat{i}+6 \hat{j}) \mathrm{m} / \mathrm{s} in a magnetic field of B⃗=2k^T.\vec{B}=2 \hat{k} T.
Then choose wrong statement.
View answer and explanation
Correct option: C
The path of the particle may be y2+z2=25y^2+z^2=25
Option C is the wrong statement. The path of the particle may be y2+z2=25y^2+z^2=25 is incorrect because the initial velocity is in the x-y plane and the magnetic field is along the z-axis, so the motion will be confined to the x-y plane.
7 Hard
7 of 14
Application +4 | -1
An insulating rod of length ll carries a charge qq distributed uniformly on it. The rod is pivoted at an end and is rotated at a frequency ff about a fixed perpendicular axis. The magnetic moment of the system is :
View answer and explanation
Correct option: D
13πqfl2\frac{1}{3} \pi q f l^2
Option D (240 years) is correct because after 4 half-lives, the ratio of the radioactive isotope to the stable element will be 1:15.
8 Hard
8 of 14
Application +4 | -1
A solenoid of 1000 turns per meter has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is (permeability of free space =  4π×10−7H/m= \,\,4\pi \times {10^{ - 7}}\mathrm{H/m} )
View answer and explanation
Correct option: A
πT\pi T
Option A (πT\pi T) is correct. The magnetic flux density is calculated using the formula B = μ_r μ_0 n I, where μ_r is the relative permeability, μ_0 is the permeability of free space, n is the number of turns per meter, and I is the current.
9 Hard
9 of 14
Application +4 | -1
An ideal coil of 10 henry is joined in series with a resistance of 5ohm5 \mathrm{ohm} and a battery of 5 volt. After two seconds joining, the current flowing in ampere in the circuit will be :-
View answer and explanation
Correct option: B
(1−e−1)\left(1-\mathrm{e}^{-1}\right)
Option B (1−e−11-\mathrm{e}^{-1}) is correct because it represents the current flowing in the L-R circuit after two seconds, calculated using the formula for current growth in an LR circuit.
10 Hard
10 of 14
Application +4 | -1
The magnitude of magnetic field at two points on the axis of a circular coil at distances of 0.05 m and 0.2 m from its centre are in the ratio 8:1.8: 1. The radius of coil is:
View answer and explanation
Correct option: B
10 cm
Option B (10 cm) is correct because it is the radius that satisfies the given ratio of magnetic fields at the specified distances from the coil's center.
11 Hard
11 of 14
Understanding +4 | -1
A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be :-
View answer and explanation
Correct option: C
A helix with uniform pitch
Option C (A helix with uniform pitch) is correct because the component of velocity parallel to the magnetic field remains constant, while the perpendicular component causes circular motion.
12 Hard
12 of 14
Application +4 | -1
There exists a uniform and constant magnetic field B=1.67TB=1.67 T along positive zz-axis. Simultaneously a proton and an alpha particle are projected from the origin with same speed v=45×105 ms−1v=4 \sqrt{5} \times 10^5 \mathrm{~ms}^{-1} along positive and negative xx-axis respectively. Find the separation between them when their velocities become mutually perpendicular for the first time. Ignore the force of interaction between them as well as any force other than magnetic force. Mass of the proton is 1.67×10−27 kg1.67 \times 10^{-27} \mathrm{~kg}.
View answer and explanation
Correct option: B
25 mm25 \mathrm{~mm}
Option C (23×104N−m\frac{2}{3} \times {10^4}N - m) is correct because the instantaneous torque is calculated by dividing the power by the angular speed.
13 Hard
13 of 14
Analyzing +4 | -1
A charged particle of unit mass and unit charge moves with velocity of v⃗=(8i^+6j^)m/s\vec{v}=(8 \hat{i}+6 \hat{j}) \mathrm{m} / \mathrm{s} in a magnetic field of B⃗=2k^T.\vec{B}=2 \hat{k} T.
Then choose wrong statement.
View answer and explanation
Correct option: C
The path of the particle may be y2+z2=25y^2+z^2=25
The correct answer is C (The path of the particle may be y2+z2=25y^2+z^2=25) because the magnetic force does no work, so the speed remains constant, and the motion is in the x-y plane.
14 Hard
14 of 14
Application +4 | -1
A galvanometer, having a resistance of 50Ω,50 \Omega, gives a full scale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross-section 2.97×102 cm22.97 \times 10^2 \mathrm{~cm}^2 that can be used to convert the galvanometer into a ammeter which can read a maximum of 5A current is (specific resistance of wire =5×10−7Ωm)=5 \times 10^{-7} \Omega \mathrm{m})
View answer and explanation
Correct option: C
3
Option C (3) is correct because it is the length of the resistance wire calculated using the shunt resistance formula for converting a galvanometer into an ammeter.
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