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NEET • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Mixed difficulty

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151 Medium
151 of 280
Application +4 | -1
A circular loop of radius 0.0157 m carries a current of 2.0 A. The magnetic field at the center of the loop is

[μ0=4π×10−7Wb−A−1m−1]\left[ {{\mathrm{\mu }_0} = 4\pi \times {{10}^{ - 7}}\mathrm{Wb} - {\mathrm{A}^{ - 1}}{\mathrm{m}^{ - 1}}} \right]
View answer and explanation
Correct option: B
8.0×10−5  Wb−m28.0 \times {10^{ - 5}}\mathrm{\;Wb} - {\mathrm{m}^2}
Option B is correct because the magnetic field at the center of a circular loop is calculated using the formula B = μ₀I / 2R.
152 Easy
152 of 280
Analyzing +4 | -1
Statement I: An electron moving in the positive xx-direction enters a region where uniform electric and magnetic fields exist perpendicular to each other. The electric field is in the negative yy-direction. If the electron travels undeflected in this region, the direction of the magnetic field is along the negative zz-axis.

Statement II: If a charged particle moves in a direction perpendicular to a magnetic field, the direction of the force acting on it is given by Fleming's left hand rule.
View answer and explanation
Correct option: B
Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I
Option C (Both vv and BB are constant) is correct because the radius of the circular path depends on both velocity and magnetic field strength, and both must be constant for the radius to remain constant.
153 Medium
153 of 280
Application +4 | -1
A straight wire of length of 0.5  metre\mathrm{0}.\mathrm{5}\,\,\mathrm{metre} and carrying a current of 1.2  ampere1.2\,\,ampere is placed in a uniform magnetic field of induction 2 tesla. If the magnetic field is perpendicular to the length of the wire, the force acting on the wire is
View answer and explanation
Correct option: B
1.2 N
Option B (1.2 N) is correct because it is calculated using the formula for the force on a current-carrying wire in a magnetic field.
154 Hard
154 of 280
Application +4 | -1
A galvanometer gives full scale deflection with 0.2 A0.2 \mathrm{~A} in its coil. The resistance of its coil is 10 Ω10\, \Omega. How much shunt resistance is required to convert it into an ammeter to read current upto 1.8 A1.8 \mathrm{~A} ?
View answer and explanation
Correct option: C
1.25 Ω1.25\, \Omega
Option C (1.25 Ω1.25\, \Omega) is correct because it's the shunt resistance value that allows the galvanometer to measure up to 1.8 A, given its internal resistance and full-scale deflection current.
155 Hard
155 of 280
Application +4 | -1
A direct current flows through a solenoid, entering in an anticlockwise sense at the end A. This solenoid, made to float on still water will:
View answer and explanation
Correct option: A
have the end A point towards north
Option D (Vsin⁡θBd\frac{{V} \sin \theta}{{Bd}}) is correct because the specific charge is calculated using the geometry of the deflection and the magnetic field.
156 Medium
156 of 280
Analyzing +4 | -1
A particle moving in the xyxy-plane experiences a velocity dependent force F=k(vyi^+vxj^),F = k\left( {{v_y}\hat i + {v_x}\hat j} \right), where vx{v_x}\mathrm{ } and vy{v_y} are the xx and yy components of its velocity v.v. If aa is the acceleration of the particle, then which of the following statements is true for the particle?
View answer and explanation
Correct option: A
Quantity v×av \times a is constant in time.
Option C (The proton trajectory will be less curved than the electron trajectory) is correct because the radius of curvature in an electric field is proportional to mass and inversely proportional to charge and electric field strength. Since the proton is much more massive, its trajectory will be less curved.
157 Medium
157 of 280
Understanding +4 | -1
A wire of length L is placed parallel to the lines
of force in a uniform magnetic field B. If I is
the current in the wire, then it will experience:
View answer and explanation
Correct option: D
no force
Option D (no force) is correct because the force on a current-carrying wire in a magnetic field is zero when the wire is parallel to the magnetic field.
158 Medium
158 of 280
Understanding +4 | -1
The magnetic dipole moment of a current carrying conducting loop depends on
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the magnetic dipole moment of a current-carrying loop is directly proportional to the number of turns, the current, and the area of the loop.
159 Hard
159 of 280
Application +4 | -1
A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 amp. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 amp. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of solenoid
View answer and explanation
Correct option: C
6×10−6Nm6 \times {10^{ - 6}}Nm
Option C (6×10−6Nm6 \times {10^{ - 6}}Nm) is correct because it is the calculated torque required to hold the coil perpendicular to the solenoid's axis.
160 Easy
160 of 280
Application +4 | -1
Charge qq is uniformly spread on a thin ring of radius RR. The ring rotates about its axis with a uniform frequency ff Hz. The magnitude of magnetic induction at the centre of the ring is
View answer and explanation
Correct option: B
μ0qf2R\frac{{{\mu _0}qf}}{{2R}}
Option B (μ0qf2R\frac{{{\mu _0}qf}}{{2R}}) is correct because the magnetic field at the center of a rotating charged ring is proportional to the charge, frequency, and inversely proportional to the radius.
161 Medium
161 of 280
Remembering +4 | -1
The cyclotron assembly is filled with
View answer and explanation
Correct option: D
Vacuum.
Option D is correct. A vacuum is used in a cyclotron to minimize collisions between the accelerated particles and air molecules.
162 Medium
162 of 280
Remembering +4 | -1
Field inside a solenoid is
View answer and explanation
Correct option: B
directly proportional to current
Option B (directly proportional to current) is correct because the magnetic field inside a solenoid is directly proportional to the current flowing through it.
163 Medium
163 of 280
Application +4 | -1
A straight wire carrying current ii is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S. unit is MM, the length of wire will be
View answer and explanation
Correct option: B
4πMi\sqrt {\frac{{4\pi M}}{i}}
Option B (4πMi\sqrt {\frac{{4\pi M}}{i}}) is correct because it correctly relates the length of the wire to the magnetic moment and current.
164 Medium
164 of 280
Analyzing +4 | -1
A charged particle is moving in the presence of electric field E⃗{\mathbf{\vec E}} and magnetic field B⃗.\mathbf{\vec B}. The directions of E⃗  \mathbf{\vec E}\; and B⃗{\mathbf{\vec B}} are such that the charged particle moves in a straight line and its speed increases. The relations amongst E⃗{\mathbf{\vec E}} , B⃗{\mathbf{\vec B}} and velocity v⃗{\mathbf{\vec v}} must be such that
View answer and explanation
Correct option: B
E⃗  ,B⃗\mathbf{\vec E}\;,\mathbf{\vec B} and v⃗{\mathbf{\vec v}} are all parallel to each other
Option B (E⃗  ,B⃗\mathbf{\vec E}\;,\mathbf{\vec B} and v⃗{\mathbf{\vec v}} are all parallel to each other) is correct because if all three vectors are parallel, the magnetic force is zero, and the electric force accelerates the particle in a straight line.
165 Medium
165 of 280
Application +4 | -1
A charged particle with charge qq enters a region of constant, uniform and mutually orthogonal fields E\mathbf{E} and B\mathbf{B} with a velocity v\mathbf{v} perpendicular to both E  and  B\mathbf{E}\;and\;\mathbf{B}, and comes out without any change in magnitude or direction of v\mathbf{v}. Then
View answer and explanation
Correct option: A
v  =  E  ×  B/B2\mathbf{v}\; = \;\mathbf{E}\; \times \;\mathbf{B}/{\mathrm{B}^2}
Option A is correct because the condition for a charged particle to move undeflected in mutually perpendicular electric and magnetic fields is v=E×B/B2\mathbf{v} = \mathbf{E} \times \mathbf{B}/{\mathrm{B}^2}.
166 Hard
166 of 280
Application +4 | -1
A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is ____.

[Take mass of electron me=9×10−31 kg,m_e=9 \times 10^{-31} \mathrm{~kg}, charge of electron ∣qe∣=1.6×10−19C,\left|{q}_{{e}}\right|=1.6 \times 10^{-19} \mathrm{C}, μ0=4π×10−7NA2,\mu_0=4 \pi \times 10^{-7} \frac{\mathrm{N}}{\mathrm{A}^2}, 1 ns=10−9 s]\left.1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right]
View answer and explanation
Correct option: A
250
The correct answer is 250 because it is the calculated number of turns per meter in the solenoid based on the given parameters and the formula relating the magnetic field to the electron's circular motion.
167 Medium
167 of 280
Understanding +4 | -1
The ratio of voltage sensitivity (VS{V_S}) and current sensitivity (Is{I_s}) of a moving coil galvanometer is
View answer and explanation
Correct option: A
1G\frac{1}{G}
Option A (1G\frac{1}{G}) is correct because voltage sensitivity is current sensitivity divided by the galvanometer resistance (G).
168 Medium
168 of 280
Understanding +4 | -1
A current carrying small loop behaves like a small magnet. If AA be its area and MM its magnetic moment, the current in the loop will be
View answer and explanation
Correct option: A
M/AM/A
Option A (M/AM/A) is correct because the current in the loop is the magnetic moment divided by the area, derived from the formula M = IA.
169 Medium
169 of 280
Application +4 | -1
The coils AA and BB having the radii in the ratio 1:21:2 carrying currents in the ratio 5:1 and have the number of turns in the ratio 1:5. The ratio of magnetic inductions at their centres is
View answer and explanation
Correct option: B
2:12:1
Option B is correct because the ratio of magnetic inductions is calculated using the formula B = (μ₀NI) / (2r), considering the ratios of radii, currents, and number of turns.
170 Medium
170 of 280
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 it represents the formula for the magnetic field at the center of a current-carrying loop.
171 Medium
171 of 280
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 (Some charges inside the wire move to the surface as a result of B{B}) is correct because the magnetic force on the moving charges inside the wire causes them to accumulate on the surface, creating a potential difference but doing no net work.
172 Easy
172 of 280
Remembering +4 | -1
The AC voltage across a resistance can be measured using a :
View answer and explanation
Correct option: A
Hot wire voltmeter
Option A (Hot wire voltmeter) is correct because hot wire voltmeters measure AC voltage based on the heating effect of the current, which is independent of the current's direction.
173 Medium
173 of 280
Application +4 | -1
A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be (1/8)th{\left( {1/8} \right)^{\mathrm{th}}} of its value at the centre of the coil, is
View answer and explanation
Correct option: B
R3R\sqrt 3
Option B is correct because the distance is calculated by setting the magnetic induction to 1/8th of its value at the center and solving for the distance using the formula for magnetic field on the axis of a circular coil.
174 Easy
174 of 280
Remembering +4 | -1
A current carrying conductor placed in a magnetic field experiences maximum force when angle between current and magnetic field is
View answer and explanation
Correct option: B
π/2\pi /2
Option B (π/2\pi /2) is correct because the force is maximum when the angle between the current and magnetic field is 90 degrees.
175 Medium
175 of 280
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 of magnetism.
176 Easy
176 of 280
Application +4 | -1
The ratio of time periods of α\alpha-particle and proton moving on circular path in uniform transverse magnetic field :-
View answer and explanation
Correct option: A
2:12: 1
Option A is correct because the time period of a charged particle in a magnetic field is proportional to its mass and inversely proportional to its charge, leading to a ratio of 2:1 for an alpha particle and a proton.
177 Medium
177 of 280
Analyzing +4 | -1
A cyclotron is used to accelerate protons (11H)\left({ }_1^1 {H}\right) deuterons (12H)\left({ }_1^2 \mathrm{H}\right) and α\alpha-particles (24He)\left({ }_2^4 \mathrm{He}\right). While exiting under similar conditions, the minimum KE\mathrm{KE} is gained by :
View answer and explanation
Correct option: C
deuterons
Option C is correct. The minimum KE is gained by deuterons because KE is proportional to (q2/m) and deuteron has the smallest q2/m ratio.
178 Medium
178 of 280
Understanding +4 | -1
In a solenoid, if number of turns is doubled. then self inductance will become
View answer and explanation
Correct option: D
quadruped.
Option D (quadruped) is correct because self-inductance is proportional to the square of the number of turns.
179 Easy
179 of 280
Remembering +4 | -1
The magnetic field BB at a point on one end of a solenoid having nn turns per metre length and carrying a current of ii ampere is given by
View answer and explanation
Correct option: B
12 μ0 ni\frac{1}{2}\,{\mu _0}\,ni
Option B (12 μ0 ni\frac{1}{2}\,{\mu _0}\,ni) is correct because it is the formula for the magnetic field at one end of a solenoid.
180 Easy
180 of 280
Understanding +4 | -1
Magnetic field dB to a current element at any
point on its axis is
View answer and explanation
Correct option: B
zero
Option B (zero) is correct because the magnetic field due to a current element at any point on its axis is zero, as the angle between dl and r is 0 or 180 degrees, making the cross product zero.
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