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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

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151 Medium
151 of 385
Application +4 | -1
Two straight parallel wires carry currents of 200   mA   and    1A200\,\,{\mkern 1mu} mA\,\,{\mkern 1mu} \mathrm{and}\,\,\,{\mkern 1mu} 1A in opposite directions. If the wires are 20 cm apart, the distance of the neutral point from the 1A1A wire is (in cm)
View answer and explanation
Correct option: D
25
Option A (5) is incorrect. Option B (15) is incorrect. Option C (20) is incorrect. Option D (25) is correct because the neutral point is located at a distance where the magnetic fields due to both wires cancel each other out. The distance from the smaller current wire is calculated, then added to the distance between the wires to find the distance from the 1A wire.
152 Medium
152 of 385
Application +4 | -1
A long cylindrical wire kept along Z{Z}-axis carries a current of density Jˉ=J0rk^\bar{J}=J_0 r \hat{k} where J0J_0 is a constant and r{r} is the radial distance from the axis of the cylinder. The magnetic induction B{B} inside the conductor at a distance d{d} from the axis of the cylinder is
View answer and explanation
Correct option: C
μ0J0d23\frac{\mu_0 J_0 d^2}{3}
Option C (μ0J0d23\frac{\mu_0 J_0 d^2}{3}) is correct because it is the magnetic induction calculated using Ampere's law and the given current density.
153 Medium
153 of 385
Application +4 | -1
A solenoid, of length 1.5 m1.5 \mathrm{~m}, has a radius of 1.5 cm1.5 \mathrm{~cm} and has a total of 1500 turns wound on it. It carries a current of 3 A3 \mathrm{~A}. Calculate the magnitude of the axial magnetic field inside the solenoid. If an electron were to move with a speed of 2×104 ms−12 \times 10^4 \mathrm{~ms}^{-1} along the axis of this current carrying solenoid, what would be the force experienced by this electron?
View answer and explanation
Correct option: A
0.38 T,00.38 \mathrm{~T}, 0
Option A (0.38 T,00.38 \mathrm{~T}, 0) is correct because the magnetic field inside the solenoid is calculated using B=μ0NIlB=\frac{\mu_0 N I}{l}, and the force on the electron moving along the axis is zero.
154 Medium
154 of 385
Understanding +4 | -1
Given the fact that
A) Magnetic field exerts force only on a moving
charge
B) Electric field exerts force on both stationary
and moving charge
C) Magnetic field exerts force on charge moving
parallel to the direction of the field.
Which of the following is true?
View answer and explanation
Correct option: B
A and B
Option B is correct because magnetic fields exert force only on moving charges, and electric fields exert force on both stationary and moving charges.
155 Medium
155 of 385
Understanding +4 | -1
Given below are two statements:

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: A
Statement I is true but Statement II is false
Option A is correct because while current sensitivity doubles with the number of turns, voltage sensitivity does not increase in the same ratio due to the increase in resistance.
156 Medium
156 of 385
Understanding +4 | -1
A charged particle moving in a magnetic field
is increased in its velocity, then its radius of the
circle
View answer and explanation
Correct option: B
increases
Option B is correct because the radius of the circular path is directly proportional to the velocity of the charged particle.
157 Medium
157 of 385
Analyzing +4 | -1
Statement-1: Net force on a current carrying loop in a non-uniform magnetic field must be non-zero.
Statement-2: Force on a current carrying wire of length dl⃗\vec{dl} placed in magnetic field B→\overrightarrow{{B}} is given by dF→=idl⃗×B→\overrightarrow{d {F}}=i \vec{dl} \times \overrightarrow{{B}}
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D is correct because Statement-1 is false, as the net force can be zero if the loop is oriented in a specific way in a non-uniform field, and Statement-2 is true, providing the correct formula for the force on a current-carrying wire.
158 Medium
158 of 385
Application +4 | -1
A particle of charge qq moves in a circle of radius rr with speed vv. Treating the circular path as a current loop with an average current, find the maximum torque exerted on the loop by a uniform field of magnitude BB.
View answer and explanation
Correct option: A
12qvrB\frac{1}{2} q v r B
Option A (12qvrB\frac{1}{2} q v r B) is correct because it's the maximum torque exerted on the loop, calculated by finding the magnetic moment and using τ=MB\tau = MB.
159 Medium
159 of 385
Application +4 | -1
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05m and 0.2m from the centre are in the ratio 8 : 1. The radius of coil is
View answer and explanation
Correct option: B
0.1 m
Option B (0.1 m) is correct because the radius of the coil is calculated using the magnetic field formula for a circular coil on its axis.
160 Medium
160 of 385
Application +4 | -1
A long straight solenoid with cross sectional radius ' aa ' and number of turns per unit length ' nn ' has a current varying width time as I(As−1)I\left(\mathrm{As}^{-1}\right). The magnitude of the electric field as a function of distance ' rr ' from the solenoid axis is
View answer and explanation
Correct option: A
nμ0a2I2r\frac{\mathrm{n} \mu_0 \mathrm{a}^2 \mathrm{I}}{2 \mathrm{r}}
Option A (nμ0a2I2r\frac{\mathrm{n} \mu_0 \mathrm{a}^2 \mathrm{I}}{2 \mathrm{r}}) is correct because the electric field outside the solenoid is derived using Faraday's law and Ampere's law.
161 Medium
161 of 385
Understanding +4 | -1
A cyclotron is used to accelerate :
View answer and explanation
Correct option: D
only positively charged particles
Option D is correct because cyclotrons are designed to accelerate positively charged particles due to the direction of the magnetic force.
162 Medium
162 of 385
Understanding +4 | -1
A current carrying circular coil, suspended
freely in a uniform external magnetic field
orients to a position of stable equilibrium. In
this state:
View answer and explanation
Correct option: A
the plane of the coil is normal to the external
magnetic field
Option A (the plane of the coil is normal to the external magnetic field) is correct because in stable equilibrium, the magnetic moment aligns with the magnetic field, making the plane of the coil perpendicular to the field.
163 Medium
163 of 385
Application +4 | -1
A wire of length 2 m2 \mathrm{~m} carrying a current of 1 A1 \mathrm{~A} is bent to form a circle, the magnetic moment of the coil is
View answer and explanation
Correct option: B
1πAm2\frac{1}{\pi} \mathrm{Am}^2
Option B (1πAm2\frac{1}{\pi} \mathrm{Am}^2) is correct because the magnetic moment is calculated using M=IAM = IA, where A is the area of the circular loop.
164 Medium
164 of 385
Application +4 | -1
For a long current carrying solenoid having inside magnetic field is 0.6 T0.6 \mathrm{~T}. Then find the magnetic energy per unit volume is
View answer and explanation
Correct option: A
1.43×105 J/m31.43 \times 10^5 \mathrm{~J} / \mathrm{m}^3
Option A (1.43×105 J/m31.43 \times 10^5 \mathrm{~J} / \mathrm{m}^3) is correct because it is the result of calculating magnetic energy per unit volume using the given magnetic field and permeability of free space.
165 Medium
165 of 385
Analyzing +4 | -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason
R.
Assertion A: For measuring the potential difference across a resistance of 600 Ω600 \,\Omega, the
voltmeter with resistance 1000 Ω1000 \,\Omega will be preferred over voltmeter with resistance 4000 Ω.4000\,\Omega.
Reason R: Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
A is not correct but R is correct
Option C (A is not correct but R is correct) is correct because a voltmeter with higher resistance is preferred, and a voltmeter with higher resistance draws smaller current.
166 Medium
166 of 385
Application +4 | -1
An electron is moving with velocity 107 m/s10^7 \mathrm{~m} / \mathrm{s} on a circular path of radius 0.57 cm0.57 \mathrm{~cm} in a magnetic field of 10−2 Wb/m210^{-2} \mathrm{~Wb} / \mathrm{m}^2. Find the value of e/m for electron in  coulomb /kg\text { coulomb } / \mathrm{kg}.
View answer and explanation
Correct option: A
1.76×10111.76 \times 10^{11}
Option A (1.76×10111.76 \times 10^{11}) is correct because the e/m ratio is calculated using the formula e/m = v/Br.
167 Medium
167 of 385
Application +4 | -1
A +6μC+6 \mu \mathrm{C} point charge is moving at a constant velocity of 8×106 ms−18 \times 10^6 \mathrm{~ms}^{-1} in the +y+y direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic field vector it produces at the following points?
(a) x=0.500 m,y=0,z=0x=0.500 \mathrm{~m}, y=0, z=0, and
(b) x=0,y=−0.500 m,z=0x=0, y=-0.500 \mathrm{~m}, z=0.
View answer and explanation
Correct option: A
−(1.92×10−5 T)k^;0-\left(1.92 \times 10^{-5} \mathrm{~T}\right) \hat{k}; 0
Option A (−(1.92×10−5 T)k^;0-(1.92 \times 10^{-5} \mathrm{~T}) \hat{{k}}; 0) is correct because it correctly calculates the magnetic field using the Biot-Savart law for a moving point charge.
168 Medium
168 of 385
Analyzing +4 | -1
Consider a wire carrying a steady current, ii placed in a uniform magnetic field B{B} perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,
View answer and explanation
Correct option: B
Some charges inside the wire move to the surface as a result of B{B}
Option B is correct because the magnetic force causes charges to redistribute within the wire, leading to a surface charge accumulation.
169 Medium
169 of 385
Understanding +4 | -1
Inside a cyclotron, a charged particle is
subjected to both an electric field and a
magnetic field. It gains kinetic energy due to
View answer and explanation
Correct option: A
only the electric field
Option A (only the electric field) is correct because the electric field accelerates the charged particle, increasing its kinetic energy, while the magnetic field only changes its direction.
170 Medium
170 of 385
Understanding +4 | -1
A straight wire of diameter 0.5 mm0.5 \mathrm{~mm} carrying a current of 1 A1 \mathrm{~A} is replaced by another wire of 1 mm1 \mathrm{~mm} diameter carrying the same current. The strength of the magnetic field far away is
View answer and explanation
Correct option: B
Same as the earlier value
Option B (Same as the earlier value) is correct because the magnetic field strength depends on the current, not the diameter of the wire.
171 Medium
171 of 385
Understanding +4 | -1
A proton is projected with a uniform velocity v along the axis of a current - carrying solenoid, then :
View answer and explanation
Correct option: D
the proton will continue to move with velocity v along the axis
Option D (the proton will continue to move with velocity v along the axis) is correct because the magnetic force on a charge moving parallel to a magnetic field is zero.
172 Medium
172 of 385
Application +4 | -1
A horizontal wire, of length 5 cm5 \mathrm{~cm} and carrying a current of 2A2\mathrm{A} placed in the middle of a long solenoid and right angles to its axis. The solenoid has 1000 turns per metre and carries a steady current I. Calculate I if the force on the wire is vertically downwards and equal to 10−4 N10^{-4} \mathrm{~N}.
View answer and explanation
Correct option: A
0.8 A0.8 \mathrm{~A}
Option A (0.8 A0.8 \mathrm{~A}) is correct because it is calculated by equating the magnetic force on the wire to the given force and solving for the current I in the solenoid.
173 Medium
173 of 385
Understanding +4 | -1
A circuit to verify Ohm’s law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit
View answer and explanation
Correct option: C
Ammeter is always connected in series and voltmeter in parallel
Option C is correct because an ammeter must be connected in series to measure current, and a voltmeter must be connected in parallel to measure potential difference.
174 Medium
174 of 385
Understanding +4 | -1
The relation between voltage sensitivity (σV)({\sigma _V}) and current sensitivity (σi)\left( {{\sigma _i}} \right) of a moving coil galvanometer is (resistance of galvanometer is G).
View answer and explanation
Correct option: A
σiG=σv\frac{{{\sigma _i}}}{G} = {\sigma _\mathrm{v}}
Option A (σiG=σv\frac{{{\sigma _i}}}{G} = {\sigma _\mathrm{v}}) is correct because voltage sensitivity is current sensitivity divided by the galvanometer resistance.
175 Medium
175 of 385
Application +4 | -1
A circular loop of radius 0.0157 m carries a current of 2 Amperes. The magnetic field at the centre of the loop is [μ∘=4π×10−7  wb/A−m] [in  wb/m2]\left[ {{\mu _ \circ } = 4\pi \times {{10}^{ - 7}}\,\,wb/A - m} \right]\,\left[ {in\,\,wb/{m^2}} \right]
View answer and explanation
Correct option: B
8×10−58 \times {10^{ - 5}}
Option B (8×10−58 \times {10^{ - 5}}) is correct because the magnetic field at the center of a circular loop is directly proportional to the current and inversely proportional to the radius.
176 Medium
176 of 385
Application +4 | -1
A long solenoid with 2000 turns per meter has a small loop of radius 3 cm3 \mathrm{~cm} placed inside the solenoid normal to its axis. If the current through the solenoid increases steadily from 1.5 A to 5.5 A5.5 \mathrm{~A} in π2100 s\frac{\pi^2}{100} \mathrm{~s}, the induced emf in the loop is
View answer and explanation
Correct option: B
0.288mV0.288 \mathrm{mV}
Option B (0.288mV0.288 \mathrm{mV}) is correct because it is the induced emf calculated using Faraday's law of induction, considering the changing magnetic field due to the solenoid's current variation.
177 Medium
177 of 385
Application +4 | -1
A particle of mass mm and charge qq enters a magnetic field BB perpendicularly with a velocity vv, the radius of the circular path described by it will be
View answer and explanation
Correct option: D
mvqB\frac{\mathrm{mv}}{\mathrm{qB}}
Option D (mvqB\frac{\mathrm{mv}}{\mathrm{qB}}) is correct because it represents the radius of the circular path derived from equating magnetic force and centripetal force.
178 Medium
178 of 385
Understanding +4 | -1
Statement-A: When a charge particle of charge 'qq' moves with a velocity vv in magnetic field of induction B⃗\vec{B} then force acting on it F⃗=q(V⃗×B⃗)\vec{F}=q(\vec{V} \times \vec{B})

Statement-B: An electron is projected in a magnetic field along the lines of force then their will be no effect on the motion of electron.
View answer and explanation
Correct option: D
Both A and B are true
Option D (Both A and B are true) is correct because Statement A correctly states the Lorentz force, and Statement B is true since the magnetic force is zero when the electron moves parallel to the magnetic field.
179 Medium
179 of 385
Application +4 | -1
A wire 28 m28 \mathrm{~m} long is bent into NN turns of circular coil of diameter 14 cm14 \mathrm{~cm} forming a solenoid of length 60 cm60 \mathrm{~cm}. Calculate the flux density inside it when a current of 5 amp5\,\mathrm{amp} passed through it. (μ0=12.57×10−7 m−1)\left(\mu_0=12.57 \times 10^{-7} \mathrm{~m}^{-1}\right)
View answer and explanation
Correct option: A
6.67×10−4 T6.67 \times 10^{-4} \mathrm{~T}
Option A (6.67×10−4 T6.67 \times 10^{-4} \mathrm{~T}) is correct because it is calculated using the formula for magnetic flux density inside a solenoid, B=μ0nIB = \mu_0 n I, where n is the number of turns per unit length.
180 Medium
180 of 385
Application +4 | -1
An electron of mass 9×10−31 kg9 \times 10^{-31} \mathrm{~kg} and charge 1.6 ×10−19C\times 10^{-19} \mathrm{C} moving with a velocity of 106 ms−110^6 \mathrm{~ms}^{-1} entered a magnetic field normally and described a circle of radius 10 cm10 \mathrm{~cm}. Then the intensity of the magnetic field is
View answer and explanation
Correct option: A
5.625×10−5 T5.625 \times 10^{-5} \mathrm{~T}
Option A (5.625×10−5 T5.625 \times 10^{-5} \mathrm{~T}) is correct because it is calculated using the formula for the magnetic force on a moving charge.
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