Magnetic Effect of Current — Page 9 | JEE Physics

Q81

The magnetic field at the center of current carrying circular loop is

B_{1}

. The magnetic field at a distance of

\sqrt{3}

times radius of the given circular loop from the center on its axis is

B_{2}

. The value of

B_{1} / B_{2}

will be

A 9 : 4

B 12 :

\sqrt5

C 8 : 1

D 5 :

\sqrt3

Correct Answer

Option C

Solution

{B_1} = {{{\mu _0}i} \over {2R}}

{B_2} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{{3 \over 2}}}}}

\Rightarrow {{{B_1}} \over {{B_2}}} = {1 \over {{R^3}}}{({R^2} + {x^2})^{{3 \over 2}}}

= {1 \over {{R^3}}}(8{R^3})

= 8

Q82

A long conducting wire having a current I flowing through it, is bent into a circular coil of

\mathrm{N}

turns. Then it is bent into a circular coil of

\mathrm{n}

turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

A

N^{2}: n^{2}

B

\mathrm{N}: \mathrm{n}

C

\mathrm{n}: \mathrm{N}

D

n^{2}: N^{2}

Correct Answer

Option A

Solution

$I=(2 \pi r) n$

\begin{aligned} & r \propto\left(\frac{I}{n}\right) \\\\ & B=n\left(\frac{\mu_{0} i}{2 r}\right) \propto\left(\frac{\mu_{0} i}{2 L}\right) n^{2} \\\\ & \frac{B_{1}}{B_{2}}=\left(\frac{N^{2}}{n^{2}}\right) \end{aligned}

Q83

A rod with circular cross-section area

2 \mathrm{~cm}^{2}

and length

40 \mathrm{~cm}

is wound uniformly with 400 turns of an insulated wire. If a current of

0.4 \mathrm{~A}

flows in the wire windings, the total magnetic flux produced inside windings is

4 \pi \times 10^{-6} \mathrm{~Wb}

. The relative permeability of the rod is (Given : Permeability of vacuum

\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}

)

A

\dfrac{5}{16}

B 125

C

\dfrac{32}{5}

D 12.5

Correct Answer

Option A

Solution

Magnetic field in the Solenoid,

\begin{aligned} & \mathrm{B}=\mu_0 \mu_r \mathrm{nI} \\\\ & \text{ Magnetic flux, } \phi=\mathrm{N}(\mathrm{BA}) \\\\ & \phi=N\left(\mu_0 \mu_r n I A\right) \\\\ & \Rightarrow 4 \pi \times 10^{-6}=400\left(4 \pi \times 10^{-7} \mu_r \times \frac{400}{0.4} \times 0.4 \times 2 \times 10^{-4}\right) \\\\ & \Rightarrow \frac{1}{40}=\mu_r \times 8 \times 10^{-2} \\\\ & \Rightarrow \mu_r=\frac{100}{320}=\frac{5}{16} \end{aligned}

Q84

The magnetic moments associated with two closely wound circular coils

\mathrm{A}

and

\mathrm{B}

of radius

\mathrm{r}_{\mathrm{A}}=10

\mathrm{cm}

and

\mathrm{r}_{\mathrm{B}}=20 \mathrm{~cm}

respectively are equal if : (Where

\mathrm{N}_{\mathrm{A}}, \mathrm{I}_{\mathrm{A}}

and

\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}

are number of turn and current of

\mathrm{A}

and

\mathrm{B}

respectively)

A

4 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}

B

2 \mathrm{~N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}

C

\mathrm{N}_{\mathrm{A}}=2 \mathrm{~N}_{\mathrm{B}}

D

\mathrm{N}_{\mathrm{A}} \mathrm{I}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}}

Correct Answer

Option D

Solution

M_A=M_B

{I_A}{N_A}\left( {\pi r_A^2} \right) = {I_B}{N_B}\left( {\pi r_B^2} \right)

{I_A}{N_A} = 4{I_B}{N_B}

Q85

For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is

4.0\times10^{-5}\mathrm{N~m~rad^{-1}}

, the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm

^2

) is :

A 1.5

B 2.0

C 0.5

D 1.0

Correct Answer

Option D

Solution

$\because \theta=\left(\dfrac{N B A}{K}\right) I$

\begin{aligned} A & =\frac{\theta K}{N B I} \\\\ & =\frac{0.05 \times 4 \times 10^{-5}}{(200) \times(0.01) \times\left(10 \times 10^{-3}\right)} \\\\ & =1 \mathrm{~cm}^{2} \end{aligned}

Q86

Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F

_1

. If distance between wires is halved and currents on them are doubled, force F

_2

on 10 cm length of wire P will be:

A

\dfrac{F_1}{8}

B 10 F

_1

C

\dfrac{F_1}{10}

D 8 F

_1

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Force per unit length between two parallel straight wires }=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{d}} \\\\ & \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\frac{\mu_0(10)^2}{2 \pi(5 \mathrm{~cm})}}{\frac{\mu_0(20)^2}{2 \pi\left(\frac{5 \mathrm{~cm}}{2}\right)}}=\frac{1}{8} \\\\ & \Rightarrow \mathrm{F}_2=8 \mathrm{~F}_1 \end{aligned}

Q87

An electron is moving along the positive

\mathrm{x}

-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then A. The electron will experience magnetic force along positive y-axis B. The electron will experience magnetic force along negative y-axis C. The electron will not experience any force in magnetic field D. The electron will continue to move along the positive

\mathrm{x}

-axis E. The electron will move along circular path in magnetic field Choose the correct answer from the options given below:

A A and E only

B B and D only

C B and E only

D C and D only

Correct Answer

Option C

Solution

The Lorentz force equation is given as:

\vec{F} = -q(\vec{v} \times \vec{B})

The electron is moving along the positive x-axis, so its velocity vector is

\vec{v} = v_x \hat{i}

. The magnetic field is applied parallel to the negative z-axis, so its magnetic field vector is

\vec{B} = -B_z \hat{k}

. Now, we can calculate the cross product of the velocity and magnetic field vectors:

\vec{v} \times \vec{B} = (v_x \hat{i}) \times (-B_z \hat{k})

Using the cross product properties, we get:

\vec{v} \times \vec{B} = -v_x B_z (\hat{i} \times \hat{k})

The cross product of

\hat{i}

and

\hat{k}

is

-\hat{j}

, so:

\vec{v} \times \vec{B} = -v_x B_z (-\hat{j}) = v_x B_z \hat{j}

Since the electron has a negative charge, the magnetic force will be in the opposite direction:

\vec{F} = -(-e)(v_x B_z \hat{j}) = e(v_x B_z \hat{j})

As a result, the electron will experience a magnetic force along the negative y-axis.

Additionally, as mentioned earlier, when a charged particle moves through a magnetic field perpendicular to its velocity, it follows a circular path.

In this case, the velocity of the electron is along the positive x-axis, and the magnetic field is along the negative z-axis, which are indeed perpendicular to each other.

As a result, the electron will move along a circular path in the magnetic field.

Hence, the correct answer is: (C) B and E only

Q88

The source of time varying magnetic field may be (A) a permanent magnet (B) an electric field changing linearly with time (C) direct current (D) a decelerating charge particle (E) an antenna fed with a digital signal Choose the correct answer from the options given below:

A (D) only

B (A) only

C (B) and (D) only

D (C) and (E) only

Correct Answer

Option A

Solution

Source of time varying magnetic field may be $\rightarrow$ accelerated or retarded charge which produces varying electric and magnetic fields.

$\rightarrow$ An electric field varying linearly with time will not produce variable magnetic field as current will be constant

Q89

An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid. A. The electron will experience magnetic force along the axis of the solenoid. B. The electron will not experience magnetic force. C. The electron will continue to move along the axis of the solenoid. D. The electron will be accelerated along the axis of the solenoid. E. The electron will follow parabolic path-inside the solenoid. Choose the correct answer from the options given below:

A B, C and D only

B B and C only

C A and D only

D B and E only

Correct Answer

Option B

Solution

The magnetic field inside a solenoid is uniform and parallel to the axis of the solenoid.

When an electron moves with constant velocity along the axis of the solenoid, the angle between its velocity vector and the magnetic field is 0°.

The magnetic force experienced by a moving charge is given by the Lorentz force formula:

\vec{F} = q(\vec{v} \times \vec{B})

Since the angle between the velocity vector and the magnetic field is 0°, the cross product term becomes zero:

\vec{v} \times \vec{B} = 0

Therefore, the magnetic force experienced by the electron is also zero:

\vec{F} = 0

As a result, the electron will not experience any magnetic force (Option B) and will continue to move along the axis of the solenoid with constant velocity (Option C).

Thus, the correct answer is: B and C only

Q90

A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be

A helical path with the axis along magnetic field

\mathrm{B}

B straight along the direction of magnetic field

\mathrm{B}

C circular path

D helical path with the axis perpendicular to the direction of magnetic field B

Correct Answer

Option A

Solution

When a charged particle moves in a magnetic field, its motion is affected by the components of its velocity that are parallel and perpendicular to the magnetic field.

The component of velocity that is parallel to the magnetic field doesn't get affected by the magnetic field.

It causes the particle to move along the magnetic field lines in a straight line.

The component of velocity that is perpendicular to the magnetic field causes the charged particle to move in a circular path around the magnetic field lines due to the magnetic Lorentz force.

Combining these two effects, the charged particle follows a helical path, where the axis of the helix is aligned with the magnetic field.