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Magnetic Effect of Current

JEE Physics · 124 questions · Page 10 of 13 · Click an option or "Show Solution" to reveal answer

Q91
A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is
A uniform in the region rar a
B zero in the region rar a
C directly proportional to rr in the region rar a
D inversely proportional to rr in the region rar a
Correct Answer
Option C
Solution

The magnetic field due to a current carrying wire can be calculated using Ampere's law.

When the current is uniformly distributed across the cross-section of the wire, the situation will be different inside and outside the wire.

Inside the wire (r < a), the magnetic field is directly proportional to r (the distance from the center of the wire).

This is because as you move away from the center of the wire, you enclose more current, so the magnetic field increases linearly with r.

Outside the wire (r > a), all the current in the wire is enclosed, so the magnetic field decreases with increasing r.

This is a result of the magnetic field lines spreading out as they move away from the wire.

Q92
A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E→\overrightarrow{\mathrm{E}} and B→\overrightarrow{\mathrm{B}} represent the electric and magnetic fields respectively, then the region of space may have : (A) E=0, B=0\mathrm{E}=0, \mathrm{~B}=0 (B) E=0, B≠0\mathrm{E}=0, \mathrm{~B} \neq 0 (C) E≠0, B=0\mathrm{E} \neq 0, \mathrm{~B}=0 (D) E≠0, B≠0\mathrm{E} \neq 0, \mathrm{~B} \neq 0 Choose the most appropriate answer from the options given below :
A (A), (B) and (C) only
B (A), (C) and (D) only
C (A), (B) and (D) only
D (B), (C) and (D) only
Correct Answer
Option C
Solution

Net force on particle must be zero i.e.

qE→+qV→×B→=0\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q} \overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=0

Possible cases are (i)

E→&B→=0\overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}=0

(ii)

V→×B→=0,E→=0\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=0, \overrightarrow{\mathrm{E}}=0

(iii)

qE→=−qV→×B→q \overrightarrow{\mathrm{E}}=-\mathrm{q} \overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}
E→≠0&B→≠0\overrightarrow{\mathrm{E}} \neq 0 \& \overrightarrow{\mathrm{B}} \neq 0
Q93
A uniform magnetic field of 2×10−3 T2 \times 10^{-3} \mathrm{~T} acts along positive YY-direction. A rectangular loop of sides 20 cm20 \mathrm{~cm} and 10 cm10 \mathrm{~cm} with current of 5 A5 \mathrm{~A} is in Y−ZY-Z plane. The current is in anticlockwise sense with reference to negative XX axis. Magnitude and direction of the torque is:
A 2×10−4 N2 \times 10^{-4} \mathrm{~N}- m\mathrm{m} along negative ZZ-direction
B 2×10−4 N2 \times 10^{-4} \mathrm{~N} - m\mathrm{m} along positive XX-direction
C 2×10−4 N2 \times 10^{-4} \mathrm{~N} - m\mathrm{m} along positive YY-direction
D 2×10−4 N2 \times 10^{-4} \mathrm{~N} - m\mathrm{m} along positive ZZ-direction
Correct Answer
Option A
Solution
M→=iA→=5×(0.2)×(0.1)(−i^)=0.1(−i^)τ⃗=M→×B→=0.1(−i^)×(2×10−3)(j^)=2×10−4(−k^)N−m\begin{aligned} & \overrightarrow{\mathrm{M}}=\mathrm{i} \overrightarrow{\mathrm{A}} \\ & =5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}}) \\ & =0.1(-\hat{\mathrm{i}}) \\ & \vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}}) \\ & =2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m} \end{aligned}
Q94
Two particles XX and YY having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii R1R_1 and R2R_2 respectively. The mass ratio of XX and YY is :
A (R1R2)\left(\dfrac{R_1}{R_2}\right)
B (R2R1)\left(\dfrac{R_2}{R_1}\right)
C (R2R1)2\left(\dfrac{R_2}{R_1}\right)^2
D (R1R2)2\left(\dfrac{R_1}{R_2}\right)^2
Correct Answer
Option D
Solution
R=mvqB=pqB=2 m(KE)qB=2mqVqBR∝mm∝R2m1 m2=(R1R2)2\begin{aligned} & \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \\ & \mathrm{R} \propto \sqrt{\mathrm{m}} \\ & \mathrm{m} \propto \mathrm{R}^2 \\ & \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2 \end{aligned}
Q95
Given below are two statements : Statement (I) : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account Statement (II) : Ampere's circuital law does not depend on Biot-Savart's law. In the light of the above statements, choose the correct answer from the options given below :
A Both Statement I and Statement II are false
B Statement I is false but Statement II is true
C Both Statement I and Statement II are true
D Statement I is true but Statement II is false
Correct Answer
Option D
Solution

Let's analyze each statement in detail to determine the correct answer: Statement (I) : "When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account."

This statement is true.

According to physics, particularly when dealing with electromagnetism and Maxwell's equations, the momentum of the electromagnetic field plays a critical role in conserving momentum in systems where electromagnetic forces are at play.

In situations where electric and magnetic fields vary with time, they can carry momentum.

Thus, for the conservation laws to hold, including Newton's third law which states that for every action, there is an equal and opposite reaction, the momentum carried by the electromagnetic fields must be included.

This is essential in scenarios such as radiation pressure where light (which can be considered an electromagnetic wave) exerts pressure on surfaces, thereby imparting momentum.

Statement (II) : "Ampere's circuital law does not depend on Biot-Savart's law."

This statement is false.

Historically and mathematically, Ampère's Circuital Law and Biot-Savart Law are closely related in the context of classical electromagnetism.

Ampère's Law, particularly in its integral form, relates the integrated magnetic field around a closed loop to the electric current passing through the loop.

Biot-Savart Law, on the other hand, is used to calculate the magnetic field generated by a current-carrying element at a point in space.

Although Ampère's Law can be derived without directly invoking the Biot-Savart Law, the fundamental understanding and derivations of magnetic fields due to currents, as presented in many textbooks and formulations, show that both laws are manifestations of how moving charges produce magnetic fields.

Moreover, the formulation of Ampère's Law was later extended by Maxwell (Maxwell's addition) to include the concept of displacement current, linking it more fundamentally to the changing electric fields and closing a conceptual loop that ties it to the broader electromagnetic theory that includes the effects described by the Biot-Savart Law.

Given the above explanations: Option D (Statement I is true but Statement II is false) is the correct answer.

Q96
An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :
A the electron will experience a force at 45∘45^{\circ} to the axis and execute a helical path.
B the electron will be accelerated along the axis.
C the electron path will be circular about the axis.
D the electron will continue to move with uniform velocity along the axis of the solenoid.
Correct Answer
Option D
Solution

For this scenario, it's important to recall how magnetic fields influence the motion of charged particles, and the configuration of the magnetic field within a solenoid.

Inside a long solenoid, the magnetic field lines run parallel to the axis of the solenoid.

The strength of this field is uniform and depends on the current running through the solenoid's coils and the number of turns per unit length but is independent of the position inside the solenoid as long as one is sufficiently far from the ends.

The force experienced by a charged particle moving in a magnetic field is given by the Lorentz force equation:

F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B})

where

qq

is the charge of the particle,

v⃗\vec{v}

is the velocity of the particle, and

B⃗\vec{B}

is the magnetic field. For an electron moving along the axis inside a solenoid: The magnetic field (

B⃗\vec{B}

) inside the solenoid is parallel to the axis of the solenoid. The velocity of the electron (

v⃗\vec{v}

) is also parallel to the axis of the solenoid and hence parallel to

B⃗\vec{B}

. Since the cross product of two parallel vectors (in this case,

v⃗\vec{v}

and

B⃗\vec{B}

) is zero, the Lorentz force (

F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B})

) acting on the electron will be zero.

Therefore, the electron will not experience any force due to the magnetic field of the solenoid, as there is no component of its velocity that is perpendicular to the magnetic field within the solenoid.

Given the above explanation, the correct option is: Option D: the electron will continue to move with uniform velocity along the axis of the solenoid.

This is because, in the absence of any force acting on it, the electron will maintain its state of motion according to Newton's first law of motion, which states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

Q97
A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a2\dfrac{a}{2} and 2a2 a from axis of the wire is :
A 4:14: 1
B 3:43: 4
C 1:11: 1
D 1:41: 4
Correct Answer
Option C
Solution

To find the ratio of the magnetic field at

a2\frac{a}{2}

and

2a2a

distances from the axis of a long straight wire, we use Ampère's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.

Ampère’s Law is given by:

∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

where:

B⃗\vec{B}

is the magnetic field,

μ0\mu_0

is the permeability of free space,

IencI_{\text{enc}}

is the enclosed current. For a point inside the wire (at radius

r=a/2r=a/2

): The current enclosed by a radius

rr

is proportional to the area of the cross-section at radius

rr

. The area of the cross-section at radius

rr

is given by:

π(a2)2=πa24\pi \left( \frac{a}{2} \right)^2 = \frac{\pi a^2}{4}

The total current

II

is uniformly distributed, thus the current enclosed

IencI_{\text{enc}}

at radius

r=a2r = \frac{a}{2}

is:

Ienc=I×Area enclosedTotal area=I×πa24πa2=I4I_{\text{enc}} = I \times \frac{\text{Area enclosed}}{\text{Total area}} = I \times \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4}

Applying Ampère’s Law inside the conductor, we get:

B⋅2π(a2)=μ0(I4)B \cdot 2 \pi \left( \frac{a}{2} \right) = \mu_0 \left( \frac{I}{4} \right)

So,

B⋅πa=μ0I4B \cdot \pi a = \frac{\mu_0 I}{4}

Therefore, the magnetic field inside the wire at

r=a2r = \frac{a}{2}

is:

Ba2=μ0I4πaB_{\frac{a}{2}} = \frac{\mu_0 I}{4 \pi a}

For a point outside the wire (at radius

r=2ar = 2a

): The total current enclosed by a radius

r=2ar = 2a

is the entire current

II

. Applying Ampère’s Law outside the conductor, we get:

B⋅2π(2a)=μ0IB \cdot 2 \pi (2a) = \mu_0 I

So,

B⋅4πa=μ0IB \cdot 4 \pi a = \mu_0 I

Therefore, the magnetic field outside the wire at

r=2ar = 2a

is:

B2a=μ0I4πaB_{2a} = \frac{\mu_0 I}{4 \pi a}

Hence, the ratio of the magnetic field at

a2\frac{a}{2}

and

2a2a

is:

Ba2B2a=μ0I4πaμ0I4πa=1:1\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1:1

So, the correct option is: Option C:

1:11:1
Q98
In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :
A inside the outer conductor
B outside the cable
C in between the two conductors
D inside the inner conductor
Correct Answer
Option B
Solution

Let's dig into how the magnetic field behaves in a coaxial cable in relation to the given options.

The principle to consider here is Ampère's law, which states that the magnetic field in a loop surrounding a current is proportional to the amount of current enclosed.

When we apply this to a coaxial cable, we must look at different regions within the cable.

Option A: Inside the outer conductor The magnetic field within the outer conductor is not zero because the current in the outer conductor itself contributes to the magnetic field in that region.

However, considering the symmetric distribution of current and the geometry of the coaxial cable, there may be a varying magnetic field within the conductor depending on the distance from the center axis.

Option B: Outside the cable Outside the coaxial cable, the net current enclosed by a path enclosing both conductors is zero because the current in the inner conductor flows in the opposite direction to the equally magnitude current in the outer conductor.

These currents being equal and opposite in direction cancel each other out, leading to a net enclosed current of zero.

According to Ampère's law, if the net enclosed current is zero, the magnetic field in that space is also zero.

Therefore, the magnetic field is zero outside the cable.

Option C: In between the two conductors In the space between the two conductors, the magnetic field is not zero.

This region only encloses the current from the inner conductor.

The magnetic field in this region is due to the current in the inner conductor and follows the right-hand rule, which would result in concentric circles of magnetic field around the inner conductor.

Since only the inner conductor's current contributes to the magnetic field in this space, Ampère's law suggests that there is a non-zero magnetic field in this region.

Option D: Inside the inner conductor Within the inner conductor, the magnetic field is not necessarily zero.

Like within the outer conductor, the magnetic field inside the inner conductor will depend on the distribution of the current within that conductor.

Utilizing the formula derived from Ampère's law for a cylindrical conductor with a uniform current distribution, the magnetic field inside the conductor increases linearly with the distance from the center axis up to the conductor's surface.

In conclusion, the correct option, based on Ampère's law and the principle that the net current enclosed determines the magnetic field outside the current's path, is: Option B: Outside the cable This is because the equal and opposite currents in the inner and outer conductors cancel each other, leading to a net magnetic field of zero outside the coaxial cable.

Q99
The electrostatic force (F1⃗)\left(\vec{F_1}\right) and magnetic force (F⃗2)\left(\vec{F}_2\right) acting on a charge qq moving with velocity vv can be written :
A F⃗1=qB⃗,F⃗2=q(B⃗×v⃗)\vec{F}_1=q \vec{B}, \vec{F}_2=q(\vec{B} \times \vec{v})
B F⃗1=qV⃗⋅E⃗,F⃗2=q(B⃗⋅V⃗)\vec{F}_1=q \vec{V} \cdot \vec{E}, \vec{F}_2=q(\vec{B} \cdot \vec{V})
C F⃗1=qE⃗,F⃗2=q(V⃗×B⃗)\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B})
D F⃗1=qE⃗,F⃗2=q(B⃗×V⃗)\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{B} \times \vec{V})
Correct Answer
Option C
Solution

The correct expressions for the electrostatic force, F⃗1\vec{F}_1, and the magnetic force, F⃗2\vec{F}_2, acting on a charge qq moving with velocity v⃗\vec{v}, are given by the Lorentz force law.

This law states that the total force acting on a charged particle in both an electric field and a magnetic field is the sum of an electrostatic force due to the electric field and a magnetic force due to the magnetic field.

The electrostatic force is given by: F⃗1=qE⃗\vec{F}_1 = q \vec{E} where E⃗\vec{E} is the electric field.

The magnetic force is given by: F⃗2=q(v⃗×B⃗)\vec{F}_2 = q(\vec{v} \times \vec{B}) where B⃗\vec{B} is the magnetic field, and ×\times denotes the cross product, indicating that the magnetic force is perpendicular both to the direction of the velocity of the charge and the direction of the magnetic field.

Therefore, the correct option is: Option C: F⃗1=qE⃗,F⃗2=q(V⃗×B⃗)\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B}) Options A, B, and D are incorrect because they do not accurately reflect the definitions of electrostatic and magnetic forces as described by the Lorentz force law.

Q100
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be :
A [a/2, 3a]
B [a/4, 3a/2]
C [a/2, 2a]
D [a/4, 2a]
Correct Answer
Option C
Solution

Maximum possible magnetic field is at the surface

Bmax⁡=μ0I2πa Bmax⁡2=μ0I4πa\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}

It can be obtained inside as well as outside the wire For inside,

μ0I4πa=μ0Ir2πa2⇒r=a2\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \\ & \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2} \end{aligned}

For outside

μ0I4πa=μ0I2πr⇒r=2a\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \mathrm{r}=2 \mathrm{a} \end{aligned}

Correct answer [a2,2a]\left[\dfrac{\mathrm{a}}{2}, 2 \mathrm{a}\right]

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