Magnetic Effects of Current - NEET Physics Chapterwise MCQs & PYQs

NEET Magnetic Effects of Current MCQs & PYQs

Practice NEET Magnetic Effects of Current Questions

Question 1:

moderate

A semi-infinite straight conductor carries a current I P is a point at perpendicular distance a
from the conductor as shown. The field at P due to the conductor is :

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Question 2:

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A point charge q is in motion with velocity \[\overrightarrow{v}] relative to an inertial axis ‘A’. The instantaneous location of q with respect to a fixed observation point P is \[\overrightarrow{r}\] as shown. \[\overrightarrow{B}\]
the magnetic field at point P is given by :

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Question 3:

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An otherwise infinite, straight wire has two concentric loops of radii a and b carrying equal
currents in opposite directions as shown. The magnetic field at the common centre is zero for

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Question 4:

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A wire loop is formed by joining two sections of radii r1 and r2 subtending an angle θ at O. The magnetic field at O is B0.

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Question 5:

moderate

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10–² weber/m². Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre is

To solve this problem, we use the formula for the magnetic field inside a long solenoid:

 

B=μ0ni,B = \mu_0 n i,

 

where:


  • BB     

    is the magnetic field at the center of the solenoid,


  • μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}     

    is the permeability of free space,


  • nn     

    is the number of turns per unit length (in meters),


  • ii     

    is the current in the solenoid.

Step 1: Magnetic Field for the First Solenoid

The first solenoid has:


  • n1=200 turns per cm=200×100=20000 turns per metern_1 = 200 \, \text{turns per cm} = 200 \times 100 = 20000 \, \text{turns per meter}     

    ,


  • i1=ii_1 = i     

    ,


  • B1=6.28×10−2 weber/m2B_1 = 6.28 \times 10^{-2} \, \text{weber/m}^2     

    .

Using the formula for

BB

, substitute

B1B_1

to find

ii

:

 

B1=μ0n1i,B_1 = \mu_0 n_1 i,

 

6.28×10−2=(4π×10−7)⋅20000⋅i.6.28 \times 10^{-2} = (4\pi \times 10^{-7}) \cdot 20000 \cdot i.

 

Simplify:

 

i=6.28×10−24π×10−7⋅20000.i = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \cdot 20000}.

 

Substitute

4π≈12.574\pi \approx 12.57

:

 

i=6.28×10−212.57×10−7⋅20000=6.28×10−22.514×10−2.i = \frac{6.28 \times 10^{-2}}{12.57 \times 10^{-7} \cdot 20000} = \frac{6.28 \times 10^{-2}}{2.514 \times 10^{-2}}.

 

i=2.5 A.i = 2.5 \, \text{A}.

 

Step 2: Magnetic Field for the Second Solenoid

The second solenoid has:


  • n2=100 turns per cm=100×100=10000 turns per metern_2 = 100 \, \text{turns per cm} = 100 \times 100 = 10000 \, \text{turns per meter}     

    ,


  • i2=i3=2.53=0.833 Ai_2 = \frac{i}{3} = \frac{2.5}{3} = 0.833 \, \text{A}     

    .

Using the formula for

BB

:

 

B2=μ0n2i2,B_2 = \mu_0 n_2 i_2,

 

Substitute the values:

 

B2=(4π×10−7)⋅10000⋅0.833.B_2 = (4\pi \times 10^{-7}) \cdot 10000 \cdot 0.833.

 

Simplify:

 

B2=4π×10−7⋅8330.B_2 = 4\pi \times 10^{-7} \cdot 8330.

 

Substitute

4π≈12.574\pi \approx 12.57

:

 

B2=12.57×10−7⋅8330=1.048×10−2.B_2 = 12.57 \times 10^{-7} \cdot 8330 = 1.048 \times 10^{-2}.

 

Final Answer:

 

B2=1.05×10−2 weber/m2\boxed{B_2 = 1.05 \times 10^{-2} \, \text{weber/m}^2}

 

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Question 6:

moderate

Two long conductors, separated by a distance d carry currents I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d. The new value of the force between them is 

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Question 7:

moderate

An electric field E and a magnetic field B applied on a proton which moves with velocity v, it goes undeflected through the region if :

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Question 8:

moderate

A particle having charge of 1 C, mass 1 kg and speed 1 m/s enters a uniform magnetic field, having magnetic induction of 1 T, at an angle θ = 30° between velocity vector and magnetic induction. The pitch of its helical path is (in meters)

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Question 9:

moderate

An electron moves in the plane of the page through two regions of space along the
dotted-line trajectory shown in the figure. There is a uniform electric field in Region-I directed
into the plane of the page (as shown). There is no electric field in Region-II. What is a
necessary direction of the magnetic field in regions I and II ? Ignore gravitational forces.

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Question 10:

moderate

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are
very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is :

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