Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 36

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Question 1
1 / -0

The force between two parallel conductors, each of length $$50\ cm$$ and distant $$20\ cm$$ apart is $$100$$ N. If the current in one conductor is double that in another one then the values will respectively be(Given $$\mu_0=4\pi\times 10^{-7}$$)

A
$$10^4A$$ and $$2\times 10^4A$$

B
$$50A$$ and $$100A$$

C
$$10A$$ and $$20 A$$

D
$$5A$$ and $$10A$$

Solution

$$F=\dfrac {\mu_0 i_1i_2 l}{2\pi r}$$
$$100=\dfrac {2\times 10^{-7}\times 2i\times 1\times \cdot 5}{\cdot 2}$$
$$10^8=i^2$$
$$i=10^4 Amp$$
Current in other wire $$=2i=2\times10^4 Amp$$
Question 2
1 / -0

The force per unit length between two long straight conductors carrying currents 3 A each in the same direction and separated by a distance of 2.0 cm is :

A
$$9\times 10^{-7}N/m$$

B
$$9\times 10^{-6}N/m$$

C
$$9\times 10^{-5}N/m$$

D
$$9\times 10^{-4}N/m$$

Solution

The force per unit length between two long straight current carrying conductors is given by $$F=\dfrac{\mu_0 I_1I_2}{2\pi d}$$
Here, current in the conductors $$I_1=I_2=3 A, $$
Distance between the two conductors, $$d=2 cm=0.02 m$$,
So, $$F=\dfrac{4\pi \times 10^{-7}\times 3\times 3}{2\pi\times 0.02}=9\times 10^{-5}N/m$$
Question 3
1 / -0

A proton, a deuteron and an $$\alpha $$ particle are accelerated through same potential difference and then they enter in a normal uniform magnetic field, the ratio of their kinetic energies will be:

A
$$2:1:3$$

B
$$1:1:2$$

C
$$1:1:1$$

D
$$1:2:4$$

Solution

Kinetic energy obtained $$ =qV $$ where $$ q $$ is the charge and $$ V $$ is the potential through which the particle is accelerated.
Also, the force due to the magnetic field is perpendicular to the direction of motion. So it cant change the speed and consequentially Kinetic energy of the particles will remain same after they enter a normal magnetic field.
In the above problem, $$ V $$ is constant.
So KE $$ \propto q $$.
The charges of the proton, deuteron and the alpha particle are in the ratio $$ 1 : 1 : 2 $$
So, their kinetic energies are in the ratio $$ 1 : 1 : 2 $$
Question 4
1 / -0

Two long straight wires of length $$l$$ lie parallel to one another and carry currents opposite to one another of magnitudes $$i_{1}$$ and $$i_{2}$$ respectively. The force experienced by each of the straight wires is ( r is the distance of their separation)

A
repulsive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l/r)$$

B
attractive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l^{2}/r)$$

C
repulsive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l^{2}/r)$$

D
attractive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l/r)$$

Solution

The flux density(B) due to current($$i_1$$) carrying conductor at distance r from it is
$$B = \dfrac{\mu_0 i_1}{2 \pi r}$$
the force experienced due to another conductor (of same length) per unit length carrying current $$i_2$$ is
$$F = B i_2 $$
$$F = \dfrac{\mu_0 i_1}{2 \pi r}i_2$$
since both conductors are of same length l, we get
$$F = \dfrac{\mu_0 i_1 i_2}{2 \pi r}l$$
Question 5
1 / -0

Two long parallel copper wires carry currents of $$5A$$ each in the opposite direction . If the wires are separated by a distance of $$0.5 m$$, then the force between the two wires is

A
$$10^{-5}N$$ attractive

B
$$10^{-5}N$$ repulsive

C
$$2\times 10^{-6}N$$ attractive

D
$$2\times 10^{-5}N$$ repulsive.

Solution

$$F=\dfrac { \mu \ { i }_{ 1 }{ i }_{ 2 } }{ 2 \pi \ d } \quad =\quad \dfrac { 2\times { 10 }^{ -7 }\times 5\times 5 }{ 0.5 } ={ 10 }^{ -5 }N$$

Since currents are opposite so the force is repulsive.
Question 6
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A charged particle, moving at right angle to a uniform magnetic field, starts moving along a circular arc of radius of curvature $$r$$. In the field it now penetrates a layer of lead and loses $${\dfrac{3}{4} }^{th}$$ of it's initial kinetic energy. The radius of curvature of it's path now will be :

A
$$4r$$

B
$$2r$$

C
$$\dfrac{r}{4}$$

D
$$\dfrac{r}{2}$$

Solution

Initially, $$r=\dfrac{mv}{qB}$$

$$r=\dfrac{\sqrt{2K_{e}m}}{qB}$$

Now $${K_{e}}'=\dfrac{1}{4}K_{e}$$

$${r}'=\dfrac{\sqrt{2\times \bigg (\dfrac{1}{4}\bigg )K_{e}m}}{qB}$$

=$$\dfrac{r}{2}$$
Question 7
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Two particles $$X$$ and $$Y$$ having equal charges, after being accelerated through the same potential differences, enter in a region of uniform magnetic field and describe circular paths of radii $$R_{1}$$ and $$R_{2}$$ respectively. The ratio of the mass of $$X$$ to that of $$Y$$ is :

A
$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{1/2}$$

B
$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{-1}$$

C
$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{2}$$

D
$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )$$

Solution

Since the particle is accelerated through $$V$$ volts
$$\because \dfrac{mv^{2}}{2}=qV $$
$$mv=\sqrt{2mqV}$$
$$R=\dfrac{mv}{qB}$$
So, $$R_1=\dfrac{\sqrt{2m_{1}qV}}{qB}$$
$$R_2=\dfrac{\sqrt{2m_{2}qV}}{qB}$$
$$\dfrac{m_1}{m_2}=\bigg (\dfrac{R_1}{R_2}\bigg )^{2}$$
Question 8
1 / -0

A long horizontal rigidly supported wire carries a current $$i_{a}=96A$$. Directly above it and parallel to it at a distance, another wire of $$0.144N$$ weight per metre is carrying a current $$i_{b}=24A$$, in a direction same as the lower wire. If the weight of the second wire is balanced by the force due to magnetic repulsion, then its distance (in mm) from the lower wire is :

A
$$9.6$$

B
$$4.8$$

C
$$3.2$$

D
$$1.6$$

Solution

Given : $$i_a = 96 \ A$$ $$i_b = 24 \ A$$
Let the separation between the wires be $$d$$.
Thus magnetic force between them $$F_m = \dfrac{\mu_o i_ai_b}{2\pi d}$$
Weight of wire $$W =0.144 \ N$$
$$\therefore$$ $$W =\dfrac{\mu_o i_a i_b}{2\pi d}$$
$$\implies \ d = \dfrac{\mu_o i_ai_b}{2\pi W}$$
$$d = \dfrac{4\pi\times 10^{-7} \times 24\times 96}{2\pi \times 0.144} =3.2\times 10^{-3} \ m= 3.2 \ mm$$
Question 9
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A proton of energy $$8eV$$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

A
$$4eV$$

B
$$2eV$$

C
$$8eV$$

D
$$6eV$$

Solution

From the formula mentioned above, momentum of particle moving in a magnetic field $$ mv=p=qBr $$
Therefore, Kinetic Energy of that particle can be written as $$ KE=\dfrac{p^2}{2m}=\dfrac{q^2B^2r^2}{2m} $$
In the same magnetic field for the same path, $$ KE \propto \dfrac{q^2}{m} $$
This ratio is same for the alpha particle and the proton. ($$ \dfrac{(2e)^2}{4 amu}=\dfrac{4e^2}{4 amu}=\dfrac{e^2}{amu} $$ ; Here $$ amu $$ is the atomic mass unit)
So, in such conditions, both will have the same energy. Hence, energy of the alpha particle will be $$ 8 eV $$ too.
Question 10
1 / -0

A deutron of kinetic energy $$50 keV$$ is describing a circular orbit of radius $$0.5 m$$ in a plane perpendicular to magnetic field $$\vec{B}$$ . The kinetic energy of the proton that describes a circular orbit of radius $$0.5 m$$ in the same plane with the same $$\vec{B}$$ is :

A
$$25 keV$$

B
$$50 keV$$

C
$$200 keV$$

D
$$100 keV$$

Solution

Deutron is $$_{1}H^{2}$$
$$r=\dfrac{mv}{qB}$$

$$r=\dfrac{\sqrt{2K.m}}{qB}$$

$$\because r_{1}=r_{2}$$
$$\dfrac{\sqrt{2\times 50\times 2}}{B}=\dfrac{\sqrt{2\times K.1}}{B}$$
$$K=100keV$$

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Moving Charges and Magnetism Test - 36

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Moving Charges and Magnetism Test - 36

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Question 1

$$10^4A$$ and $$2\times 10^4A$$

$$50A$$ and $$100A$$

$$10A$$ and $$20 A$$

$$5A$$ and $$10A$$

Solution

Question 2

$$9\times 10^{-7}N/m$$

$$9\times 10^{-6}N/m$$

$$9\times 10^{-5}N/m$$

$$9\times 10^{-4}N/m$$

Solution

Question 3

$$2:1:3$$

$$1:1:2$$

$$1:1:1$$

$$1:2:4$$

Solution

Question 4

repulsive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l/r)$$

attractive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l^{2}/r)$$

repulsive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l^{2}/r)$$

attractive and equal to$$(\mu _{0}/2\pi )(i_{1}i_{2}l/r)$$

Solution

Question 5

$$10^{-5}N$$ attractive

$$10^{-5}N$$ repulsive

$$2\times 10^{-6}N$$ attractive

$$2\times 10^{-5}N$$ repulsive.

Solution

Question 6

$$4r$$

$$2r$$

$$\dfrac{r}{4}$$

$$\dfrac{r}{2}$$

Solution

Question 7

$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{1/2}$$

$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{-1}$$

$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )^{2}$$

$$\bigg (\dfrac{R_{1}}{R_{2}}\bigg )$$

Solution

Question 8

$$9.6$$

$$4.8$$

$$3.2$$

$$1.6$$

Solution

Question 9

$$4eV$$

$$2eV$$

$$8eV$$

$$6eV$$

Solution

Question 10

$$25 keV$$

$$50 keV$$

$$200 keV$$

$$100 keV$$

Solution

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