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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

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Showing 181-210 of 385 questions

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181 Medium
181 of 385
Application +4 | -1
When a proton has a velocity v⃗=(2i^+3j^)×106 ms−1\vec{v}=(2 \hat{i}+3 \hat{j}) \times 10^6 \mathrm{~ms}^{-1}, it experiences a force F⃗=−(1.28×10−13k^)N\vec{F}=-\left(1.28 \times 10^{-13} \hat{k}\right) \mathrm{N}. When its velocity is along the zz-axis, it experiences a force along the xx-axis. What is the magnetic field?
View answer and explanation
Correct option: A
−(0.4j^)T-(0.4 \hat{j}) \mathrm{T}
Option A (−(0.4j^)T-(0.4 \hat{j}) \mathrm{T}) is correct because it is found by applying the Lorentz force equation and analyzing the force directions for the given velocities.
182 Medium
182 of 385
Analyzing +4 | -1
Statement I - A bar magnet can exert torque on itself due to it's own field.

Statement II - One element of a current carrying wire can exert a force on another element of same wire.
View answer and explanation
Correct option: A
Statement II is true; Statement II is true
Option A is correct because a bar magnet cannot exert a torque on itself due to its own field, and one element of a current-carrying wire can exert a force on another element of the same wire.
183 Medium
183 of 385
Application +4 | -1
A current of 4 A4 \mathrm{~A} produces a deflection of 30∘30^{\circ} in the galvanometer. The figure of merit is
View answer and explanation
Correct option: B
7.6 A/rad7.6 \mathrm{~A} / \mathrm{rad}
Option B (7.6 A/rad) is correct because the figure of merit is correctly calculated using the given current and deflection.
184 Medium
184 of 385
Understanding +4 | -1
A magnetic field can be produced by :-
View answer and explanation
Correct option: D
Both (1) &\& (2)
Option D (Both (1) & (2)) is correct because both a moving charge and a changing electric field can produce a magnetic field.
185 Medium
185 of 385
Application +4 | -1
A proton carrying 1 MeV kinetic energy is moving
in a circular path of radius RR in uniform magnetic
field. What should be the energy of an a-particle
to describe a circle of the same radius in the same
field?
View answer and explanation
Correct option: B
1 MeV
Option B (1 MeV) is correct because the energy of the alpha particle is equal to the energy of the proton when they have the same radius in the same magnetic field.
186 Medium
186 of 385
Analyzing +4 | -1
Assertion (A) : A charged particle moves along positive yy-axis with constant velocity in uniform electric and magnetic fields. If magnetic field is acting along positive xx-axis, then electric field should act along positive zz-axis.

Reason (R) : To keep the charged particle undeviated the relation E=B×v\mathbf{E}=\mathbf{B} \times \mathbf{v} must hold good.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
Option A (If both Assertion and Reason are true and the Reason is correct explanation of the Assertion) is correct because the electric and magnetic forces must balance to keep the charged particle undeviated, and the given relation ensures this.
187 Medium
187 of 385
Understanding +4 | -1
In a moving coil galvanometer, we use a radial magnetic feild so that the galvanometer scale is
View answer and explanation
Correct option: C
Linear
The correct answer is Linear because a radial magnetic field ensures that the torque on the coil is directly proportional to the current, resulting in a linear scale.
188 Medium
188 of 385
Understanding +4 | -1
To which of the following quantities, the radius
of the circular path of a charged particle
moving at right angles to a uniform magnetic
field is directly proportional?
View answer and explanation
Correct option: D
momentum of the particle
Option D is correct because the radius of the circular path of a charged particle in a magnetic field is directly proportional to the momentum of the particle.
189 Medium
189 of 385
Understanding +4 | -1
Two galvanometers ' G1G_1 ' and ' G2G_2 ' require 2 mA2 \mathrm{~mA} and 3 mA3 \mathrm{~mA} respectively to produce the same deflection. Then
View answer and explanation
Correct option: D
G1G_1 is more sensitive than G2G_2
Option D (G1G_1 is more sensitive than G2G_2) is correct because sensitivity is inversely proportional to the current required for the same deflection. Since G1G_1 requires less current, it is more sensitive.
190 Medium
190 of 385
Understanding +4 | -1
A charged particle moves through a magnetic field perpendicular to its direction, then
View answer and explanation
Correct option: C
Momentum changes but kinetic energy constant
Option C (Momentum changes but kinetic energy constant) is correct because the magnetic force only changes the direction of the velocity, not its magnitude.
191 Medium
191 of 385
Application +4 | -1
A magnetic dipole is placed under the effect of two magnetic fields inclined at 75∘75^{\circ} to each other. One of the field has a magnetic induction of 1.5×10−2 T1.5 \times 10^{-2} \mathrm{~T}. The magnet comes to rest at an angle of 30∘30^{\circ} with the direction of this field. The magnitude of the other field is
View answer and explanation
Correct option: C
1.52×10−2 T\frac{1.5}{\sqrt{2}} \times 10^{-2} \mathrm{~T}
The correct answer is C (1.52×10−2 T\frac{1.5}{\sqrt{2}} \times 10^{-2} \text{{ T}}) because the magnetic dipole aligns where the torques due to both fields balance each other.
192 Medium
192 of 385
Understanding +4 | -1
A charge particle moving in magnetic field BB, has the components of velocity along BB as well as perpendicular to BB. The path of the charge particle will be
View answer and explanation
Correct option: B
Helical path with the axis along magnetic field BB
Option B (Helical path with the axis along magnetic field BB) is correct because the component of velocity along the magnetic field causes linear motion, while the perpendicular component causes circular motion, resulting in a helical path.
193 Medium
193 of 385
Understanding +4 | -1
Identify the correct statement from the following
View answer and explanation
Correct option: C
Cyclotron frequency is independent of speed of charged particle
Option C (Cyclotron frequency is independent of speed of charged particle) is correct because the cyclotron frequency depends only on the magnetic field and the charge-to-mass ratio.
194 Medium
194 of 385
Understanding +4 | -1
A long solenoid of length L has a mean
diameter D. It has n layers of windings of N
turns each. If it carries a current I, the
magnetic field at its centre will be
View answer and explanation
Correct option: C
independent of D
Option C (independent of D) is correct because the magnetic field at the center of a long solenoid depends on the number of turns per unit length (n) and the current (I), but not on the diameter (D).
195 Medium
195 of 385
Remembering +4 | -1
Cyclotron is a device which is used to:
View answer and explanation
Correct option: C
Accelerate protons
Option C (Accelerate protons) is correct because a cyclotron is designed to accelerate heavy charged particles like protons.
196 Medium
196 of 385
Analyzing +4 | -1
Assertion: Magnetic force between two charges is generally much smaller than the electric force between them.
Reason: Speeds of charges are much smaller than the free–space speed of light.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the magnetic force is much smaller than the electric force due to the relatively low speeds of charges compared to the speed of light.
197 Medium
197 of 385
Application +4 | -1
A proton enters into a magnetic field of induction 1.732 T1.732 \mathrm{~T}, with a velocity of 107 m/s10^7 \mathrm{~m} / \mathrm{s} at an angle 60∘60^{\circ} to the field. The force acting on the proton is (e=1.6×10−19C,sin⁡60∘=cos⁡30∘=\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \sin 60^{\circ}=\cos 30^{\circ}=\right. 32\frac{\sqrt{3}}{2} )
View answer and explanation
Correct option: D
2.4×10−12 N2.4 \times 10^{-12} \mathrm{~N}
Option D (2.4×10−12 N2.4 \times 10^{-12} \mathrm{~N}) is correct because the force acting on the proton is calculated using the formula F=qvBsin⁡θF = qvB \sin{\theta}.
198 Medium
198 of 385
Application +4 | -1
A long straight wire carries a current 15 A15 \mathrm{~A}. Calculate the magnetising field H\mathrm{H} at a point at distance 0.105 m0.105 \mathrm{~m} from the axis of wire
View answer and explanation
Correct option: C
22.7 A/m22.7 \mathrm{~A} / \mathrm{m}
Option C (22.7 A/m) is correct because it is calculated using the formula for the magnetic field around a long straight wire.
199 Medium
199 of 385
Application +4 | -1
A circular coil carrying a certain current produces a magnetic field B0B_0 at its centre. The coil is now rewound so as to have 4 turns and the same current is passed through it. The new magnetic field at the centre is
View answer and explanation
Correct option: B
16B016 B_0
Option A (245.01 V/m245.01 \mathrm{~V} / \mathrm{m}) is correct because it is the amplitude of the electric field of radiation at the given distance from the lamp, calculated using the Poynting vector and the relationship between electric and magnetic fields.
200 Medium
200 of 385
Analyzing +4 | -1
Statement-I: If by mistake a voltmeter is connected in series it gets burnt
Statement-II: Current will drastically decrease in the circuit.
View answer and explanation
Correct option: D
Statement-I is false, but Statement-II is true
Option D (Statement-I is false, but Statement-II is true) is correct because a voltmeter has high resistance, so connecting it in series drastically decreases the current, but it doesn't necessarily burn out.
201 Medium
201 of 385
Analyzing +4 | -1
The turns of a solenoid, designed to provide a given magnetic
flux density along its axis, are wound to fill the space between
two concentric cylinders of fixed radii. How should the diameter
dd of the wire used be chosen so as to minimize the heat
dissipated in the windings?
View answer and explanation
Correct option: C
Wire should be multiple of dd
Option C (Wire should be multiple of dd) is correct because the heat dissipated is independent of the diameter of the wire, so any multiple of d will work, but the most practical is a multiple of d itself.
202 Medium
202 of 385
Understanding +4 | -1
Fleming's left and right hand rules are used in:
View answer and explanation
Correct option: C
DC motor and DC generator
Option C (DC motor and DC generator) is correct because Fleming's left-hand rule is used for DC motors, and Fleming's right-hand rule is used for DC generators.
203 Medium
203 of 385
Remembering +4 | -1
The magnetic induction at the centre of a current carrying circle of radius rr, is
View answer and explanation
Correct option: B
Inversely proportional to rr
Option B (Inversely proportional to rr) is correct because the magnetic field at the center of a current-carrying circle is inversely proportional to the radius.
204 Medium
204 of 385
Application +4 | -1
A charge of 3.14×10−6C3.14 \times 10^{-6} \mathrm{C} is distributed uniformly over a circular ring of radius 20.0 cm20.0 \mathrm{~cm}. The ring rotates about its axis with an angular velocity of 60.0 rad s−160.0\, \mathrm{rad}\, \mathrm{s}^{-1}. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm5.00 \mathrm{~cm} from the centre.
View answer and explanation
Correct option: D
38\frac{3}{8}
Option D (38\frac{3}{8}) is correct because it is the ratio of the electric field to the magnetic field calculated using the given parameters and formulas.
205 Medium
205 of 385
Understanding +4 | -1
Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert:
View answer and explanation
Correct option: A
an attractive force on each other
Option A (an attractive force on each other) is correct because parallel currents attract each other due to the magnetic fields they create.
206 Medium
206 of 385
Application +4 | -1
A solenoid 60 cm60 \mathrm{~cm} long and of radius 4.0 cm4.0 \mathrm{~cm} has 3 layers of windings of 300 turns each. A 2.0 cm2.0 \mathrm{~cm} long wire of mass 2.5 g2.5 \mathrm{~g} lies inside the solenoid near its centre normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A6.0 \mathrm{~A} in the wire. What value of current in the windings of the solenoid can support the weight of the wire ? g=9.8 ms−2g=9.8 \mathrm{~ms}^{-2}.
View answer and explanation
Correct option: A
108.3 A
Option A (108.3 A) is correct because the current required to support the wire is calculated by equating the magnetic force on the wire to its weight.
207 Medium
207 of 385
Application +4 | -1
An electron having charge ' ee ' is moving with a constant speed vv along a circle of radius rr. Its magnetic moment will be
View answer and explanation
Correct option: B
evr2\frac{\mathrm{e} v \mathrm{r}}{2}
Option B (evr2\frac{\mathrm{e} v \mathrm{r}}{2}) is correct because the magnetic moment is proportional to the current and the area of the loop.
208 Medium
208 of 385
Application +4 | -1
A 0.8 m long solenoid has 800 turns and has a field density of 2.52×10−3T2.52\mathrm{ } \times \mathrm{ }{10^{ - 3}}T at its centre. What is current in the wire ?
View answer and explanation
Correct option: B
2
Option B (2) is correct because the current is calculated using the formula for the magnetic field inside a solenoid.
209 Medium
209 of 385
Application +4 | -1
A galvanometer of resistance 25  Ω25\,\,\mathrm{\Omega } giving full scale deflection for a current of 10  milli  ampere10\,\,milli\,\,ampere, is to be changed into a voltmeter of range 100  V100\,\,V by connecting a resistance of ′R′'R' in series with galvanometer. The value of resistance RR in Ω\mathrm{\Omega } is
View answer and explanation
Correct option: D
9975
Option D (9975) is correct because the series resistance needed to convert a galvanometer into a voltmeter is calculated using the formula R = V/ig - G.
210 Medium
210 of 385
Application +4 | -1
A galvanometer has resistance 500 ohm. It is shunted so that its sensitivity decreases by 100 times. Find the shunt resistance.
View answer and explanation
Correct option: C
5.05 Ω5.05\,\Omega
Option C (5.05 Ω5.05\,\Omega) is correct because the shunt resistance is calculated using the formula S=Gn−1S = \frac{G}{{n - 1}}, where G is the galvanometer resistance and n is the factor by which sensitivity decreases.
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