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:

difficult

A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic
vector field \[\overrightarrow{B}\] at a point having coordinates (x, y) in the z = 0 plane is :

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Question 2:

difficult

A coaxial cable having radius “a” of inner wire and inner and outer radii “b” and “c” respectively
of the outer shell carries equal and opposite currents of magnitude i on the inner and outer
conductors as shown. What is the magnitude of the magnetic induction at point P of the cable at
a distance r (b < r < c) from the axis?

To calculate the magnetic field

BB

at a point

PP

within the outer shell of a coaxial cable (

b<r<cb < r < c

), carrying equal and opposite currents

ii

on the inner and outer conductors, we use Ampère's law and superposition principles.

1. Current Distribution in the Outer Shell

The outer shell carries current

−i-i

, distributed uniformly across the cross-sectional area of the shell between radii

bb

and

cc

.

The current density

JJ

in the shell is:

 

J=−iπ(c2−b2).J = \frac{-i}{\pi (c^2 - b^2)}.

 

The current enclosed within a radius

rr

(

b<r<cb < r < c

) in the outer shell is the current

−i-i

contributed by the region from

bb

to

rr

:

 

Ienc,shell=J⋅area of shell portion=J⋅π(r2−b2).I_{\text{enc,shell}} = J \cdot \text{area of shell portion} = J \cdot \pi (r^2 - b^2).

 

Substituting

JJ

:

 

Ienc,shell=−iπ(c2−b2)⋅π(r2−b2)=−i(r2−b2)c2−b2.I_{\text{enc,shell}} = \frac{-i}{\pi (c^2 - b^2)} \cdot \pi (r^2 - b^2) = \frac{-i (r^2 - b^2)}{c^2 - b^2}.

 

2. Net Enclosed Current at Radius rr

 

At any point

PP

within the shell (

b<r<cb < r < c

), the net current enclosed by a loop of radius

rr

is:

 

Ienc=current from the inner wire+current from the outer shell.I_{\text{enc}} = \text{current from the inner wire} + \text{current from the outer shell}.

 

The inner wire contributes

+i+i

, and the shell contributes

Ienc,shellI_{\text{enc,shell}}

:

 

Ienc=i+−i(r2−b2)c2−b2.I_{\text{enc}} = i + \frac{-i (r^2 - b^2)}{c^2 - b^2}.

 

Simplify:

 

Ienc=i(1−r2−b2c2−b2).I_{\text{enc}} = i \left(1 - \frac{r^2 - b^2}{c^2 - b^2}\right).

 

Factorize:

 

Ienc=i⋅c2−r2c2−b2.I_{\text{enc}} = i \cdot \frac{c^2 - r^2}{c^2 - b^2}.

 

3. Magnetic Field at Radius rr

 

Using Ampère's law, the magnetic field

BB

at radius

rr

is:

 

B⋅2πr=μ0Ienc.B \cdot 2\pi r = \mu_0 I_{\text{enc}}.

 

Substitute

IencI_{\text{enc}}

:

 

B⋅2πr=μ0⋅i⋅c2−r2c2−b2.B \cdot 2\pi r = \mu_0 \cdot i \cdot \frac{c^2 - r^2}{c^2 - b^2}.

 

Solve for

BB

:

 

B=μ0i2πr⋅c2−r2c2−b2.B = \frac{\mu_0 i}{2\pi r} \cdot \frac{c^2 - r^2}{c^2 - b^2}.

 

Final Answer:

 

B=μ0i2πr⋅c2−r2c2−b2\boxed{B = \frac{\mu_0 i}{2\pi r} \cdot \frac{c^2 - r^2}{c^2 - b^2}}

 

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Question 3:

difficult

An electron \left(mass = 9.1\times 10^{-31}kg; Charge=-1.6\times 10^{-19}C \right) experiences no deflection if subjected to an electric field of \[3.2\times 10^{5} V/m\] and a magnetic field of 2.0 × 10-³ Wb/m² . Both the fields are normal to the path of electron and to each other . If the electric field is removed, then the electron will revolve in an orbit of radius:

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Question 4:

difficult

A long wire carrying a current of 2 A is laid along the x axis (current flows along positive
x direction) and another wire carrying current of 4 A is laid along y axis(current flows along
positive y direction). The points at which magnetic field is zero are:

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Question 5:

difficult

Three rings, each having equal radius R, are placed mutually perpendicular to each other
and each having its centre at the origin of coordinate system. If current I is flowing thriugh
each ring then the magnitude of the magnetic field at the common centre is

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Question 6:

difficult

In the given figure the magnitude of magnetic field at O is (all three wires are quarter circular arc)

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Question 7:

difficult

A current I flows around a closed path in the horizontal plane of the circle as shown in the figure. The path consists of eight arcs with alternating radii r and 2r. Each segment of arc subtends equal angle at the common centre P. The magnetic field produced by current path at point P is

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Question 8:

difficult

A plastic disc of radius R has a charge q uniformly distributed over its surface. If the disc is rotated with a frequency f about its axis. then magnetic dipole moment will be :

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Question 9:

difficult

A cube made of wires of equal length is connected to a battery as shown in the figure. The magnetic field at the centre of the cube is :

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Question 10:

difficult

Adjoining figure shows a rectangular loop of conductor carrying a current i. The length and breadth of the loop are respectively a and b. The magnetic field at the centre of loop is :

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