Magnetic Effects of Current - NEET Physics Questions
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Magnetic Effects of Current

Practice 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 :

1. \[\frac{\mu_{0}I}{2\pi}\left( \frac{y\hat{i}-x\hat{j}}{x^{2}+y^{2}} \right)\]
2. \[\frac{\mu_{0}I}{2\pi}\left( \frac{x\hat{i}+y\hat{j}}{x^{2}+y^{2}} \right)\]
3. \[\frac{\mu_{0}I}{2\pi}\left( \frac{x\hat{j}-y\hat{i}}{x^{2}+y^{2}} \right)\]
4. \[\frac{\mu_{0}I}{2\pi}\left( \frac{x\hat{i}-y\hat{j}}{x^{2}+y^{2}} \right)\]
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Question 2: 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 :

1. \[B=\frac{\mu_{0}I}{4\pi a}(1+sin\phi):outwards\]
2. \[B=\frac{\mu_{0}I}{4\pi a}(1+cos\phi):outwards\]
3. \[B=\frac{\mu_{0}I}{4\pi a}(1+sin\phi):inwards\]
4. \[B=\frac{\mu_{0}I}{4\pi a}(1+cos\phi):inwards\]
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Question 3: moderate

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 :

1. \[\overrightarrow{B}=\frac{\mu_{0}q}{4\pi}\frac{(\overrightarrow{r}\times \overrightarrow{v})}{r^{3}}\]
2. \[\overrightarrow{B}=\frac{\mu_{0}q}{2\pi}\frac{(\overrightarrow{v}\times \overrightarrow{r})}{r^{3}}\]
3. \[\overrightarrow{B}=\frac{\mu_{0}q}{4\pi}\frac{(\overrightarrow{v}\times \overrightarrow{r})}{r^{3}}\]
4. Zero
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Question 4: moderate

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

1. \[\frac{a}{b}=\frac{\pi+1}{\pi}\]
2. \[\frac{a}{b}=\frac{\pi}{\pi+1}\]
3. \[\frac{a}{b}=\frac{\pi-1}{\pi+1}\]
4. \[\frac{a}{b}=\frac{\pi+1}{\pi-1}\]
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Question 5: moderate

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.

1. \[B_{0}=\frac{\mu_{0}I}{4\pi}\left( \frac{1}{r_{1}}+\frac{1}{r_{2}} \right)\theta : outwards\]
2. \[B_{0}=\frac{\mu_{0}I}{4\pi}\left( \frac{1}{r_{1}}-\frac{1}{r_{2}} \right)\theta : inwards\]
3. \[B_{0}=\frac{\mu_{0}I}{2\pi}\left( \frac{1}{r_{1}}+\frac{1}{r_{2}} \right)\theta : outwards\]
4. \[B_{0}=\frac{\mu_{0}I}{2\pi}\left( \frac{1}{r_{1}}-\frac{1}{r_{2}} \right)\theta : inwards\]
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Question 6: 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

1. \[1.05\times 10^{-2} weber/m^{2}\]
2. \[1.05\times 10^{-5} weber/m^{2}\]
3. \[1.05\times 10^{-3} weber/m^{2}\]
4. \[1.05\times 10^{-4} weber/m^{2}\]
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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 7: 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?

1. Zero
2. \[\frac{\mu_{0}ir}{2\pi a^{2}}\]
3. \[\frac{\mu_{0}i}{2\pi r}\]
4. \[\frac{\mu_{0}i}{2\pi r}\frac{c^{2}-r^{2}}{c^{2}-b^{2}}\]
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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 8: easy

Following figures show the arrangement of bar magnets in different configuration. Each
magnet has magnetic dipole moment \[\overrightarrow{m}\] . Which configuration has highest net magnetic dipole moment?

1.
2.
3.
4.
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Question 9: 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 

1. – 2F
2. F/3
3. – 2F/3
4. – F/3
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Question 10: 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 :

1. E ⊥ B
2. E is parallel to v and perpendicular to B
3. E, B and v all three mutually perpendicular to each other and v = E/B
4. E and B both are parallel but perpendicular to v
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