Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 72

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Question 1
1 / -0

The magnitude of force per unit length on a wire carrying current $$I$$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $$L$$ and current in them is $$I$$) is:

A
$$\dfrac{+\mu_{0}I^{2}}{\pi L}\hat{j}$$

B
$$\dfrac{-\mu_{0}I^{2}}{L}\hat{j}$$

C
$$\dfrac{\mu_{0}I^{2}}{2L}\hat{j}$$

D
$$\dfrac{\mu_{0}I^{2}}{L}\hat{j}$$

Solution

Net Magnetic field at point O due to both the wires shall be
$$\displaystyle B=-2\int_{90^{0}}^{0^{0}}\frac{\mu_0 I}{2\pi l} \sin\theta d\theta$$ along $$-k$$

$$\displaystyle \therefore B= -\dfrac{\mu_0 I}{\pi l}$$ along $$-k$$

Now, $$F=I(\vec{dl} \times \vec{B})$$
Where, $$\vec{dl}$$ is along $$i$$
Therefore, $$\dfrac{F}{dl}= \dfrac{\mu I^{2}}{\pi l}$$ along $$j$$
Question 2
1 / -0

A charged particle having charge q experiences a force $$\vec{F}=q(-\vec{j}+\vec{k})N$$ in a magnetic field B when it has a velocity $$v_{1} = 1\hat i \ m/s$$. The force becomes $$\vec{F}=q(\vec{i}-\vec{k})N$$ when the velocity is changed to $$v_{2}=1\hat{j}m/sec$$. The magnetic induction vector at that point is :

A
$$(\hat{i}+\hat{j}+\hat{k})\ T$$

B
$$(\hat{i}-\hat{j}-\hat{k})\ T$$

C
$$(-\hat{i}-\hat{j}+\hat{k})\ T$$

D
$$(\hat{i}+\hat{j}-\hat{k})\ T$$

Solution

Let the magnetic induction at required point be
$$\bar { B } =\left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) T$$
Magnetic Force $$\left( \bar { F } \right) $$ on charged particle (q) moving with Velocity (V) in magnetic field $$\left( \bar { B } \right) $$ is given as
$$\bar { F } =q\left( \bar { V } \times \bar { B } \right) $$
Case 1: $$q\left( -\hat { j } +\hat { k } \right) =q\left( \hat { i } \times \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) $$
$$-\hat { j } +\hat { k } ={ B }_{ y }\hat { k } -{ B }_{ z }\hat { j } $$
$${ B }_{ z }=1....\left( i \right) $$
$${ B }_{ y }=+1....\left( ii \right) $$
Case 2: $$q\left( \hat { j } -\hat { k } \right) =q\left( \hat { j } \left( { B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \right) \right) $$
$$\therefore \hat { j } -\hat { k } =-{ B }_{ x }\hat { k } +{ B }_{ z }\hat { i } $$
$$\therefore { B }_{ x }=1....\left( iii \right) $$
From $$\left( i \right) ,\left( ii \right) \& \left( iii \right) $$
$$\bar { B } =\left( \hat { i } +\hat { j } +\hat { k } \right) T$$ Thus, option A
Question 3
1 / -0

A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

A
2.5$$\mu$$T

B
5.0$$\mu$$T

C
2.0$$\mu$$T

D
None

Solution

$$\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { q\overrightarrow { v } \times \overrightarrow { r } }{ { r }^{ 3 } } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { qvsin\theta }{ { r }^{ 2 } } $$
=$$\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { 2\times 100\times sin{ 30 }^{ 0 } }{ 4 } $$
=$${ 10 }^{ -7 }\times 25$$
Question 4
1 / -0

An $$\alpha $$ particle is moving along a circle of radius $$R$$ with a constant angular velocity $$\omega.$$ Point $$A$$ lies in the same plane at a distance $$2R$$ from the centre. Point $$A$$ records magnetic field produced by $$\alpha $$ particle. If the minimum time interval between two successive times at which $$A$$ records zero magnetic field is $$'t'$$, the angular speed $$\omega ,$$ in terms of $$t$$ is

A
$$\displaystyle \frac{2\pi }{t}$$

B
$$\displaystyle \frac{2\pi }{3t}$$

C
$$\displaystyle \frac{\pi }{3t}$$

D
$$\displaystyle \frac{\pi }{t}$$

Solution

Magnetic field using Biot Savart law:
$$\vec{B}= \dfrac{\mu_0}{4\pi}i\dfrac{(\vec{dl}\times \vec{r})}{\vec{r}^3}$$

Though the expression above has current i, it can be recast into
$$\vec {B}= \dfrac{\mu_0}{4\pi}q\dfrac{(\vec{v}\times \vec{r})}{\vec{r}^3}$$

Point A shall record zero magnetic field (due to $$\alpha$$ particle) when
the -particle is at position P and Q as shown in figure since $$\vec{v} \parallel \vec{r}$$

For any other position of the $$\alpha$$ particle $$\vec{v} $$ is not parallel $$\vec{r}$$
$$\therefore \vec{v}\times \vec{r}$$ is not equal to zero, magnetic field is not equal to zero.
$$\angle POQ =120^0$$
$$ \therefore \omega t= 120^0=\dfrac{2\pi}{3}$$
$$ \Rightarrow \omega = \dfrac{2 \pi}{3t}$$
Question 5
1 / -0

A charged particle $$A$$ of charge $$q = 2\ C$$ has velocity $$v = 100 \ m/s$$. When it passes through point $$A$$ and has velocity in the direction shown, the strength of magnetic field at point $$B$$ due to this moving charge is $$(r = 2\ m)$$

A
$$ 2.5\ \mu T$$

B
$$ 5.0\ \mu T $$

C
$$ 2.0\ \mu T$$

D
$$none\ of\ these$$

Solution

Magnetic field using Biot Savart law
$$\vec{B}= \dfrac{\mu_0}{4\pi}i\dfrac{(\vec{dl}\times \vec{r})}{\vec{r}^3}$$
Though the expression above has current $$i$$, it can be recast into
$$\vec {B}= \dfrac{\mu_0}{4\pi}q\dfrac{(\vec{v}\times \vec{r})}{\vec{r}^3}$$
$$ \therefore \left |\vec {B} \right | =( 10^{-7} ) \times 2 \times \dfrac{ 100 \times ( \sin \theta)}{2^2}$$
$$ \therefore \left |\vec {B} \right | =( 10^{-7} ) \times 2 \times \dfrac{ 100 \times \sin 30^o}{4}$$
$$ \therefore \left |\vec {B} \right | = 2.5 \times 10^{-6} \: T = 2.5 \: \mu T$$
Question 6
1 / -0

A charged particle moves through a magnetic field in a direction perpendicular to it. Then the

A
Speed of the particle remains unchanged

B
Direction of the particle remains unchanged

C
Acceleration remains unchanged

D
Velocity remains unchanged

Solution

[]We know that $$Force $$ on a charged prtice $$q$$ due to magnetic field $$\vec{B}$$ is given by

$$\vec{F}=q(\vec{V}\times\vec{B})$$ where $$\vec{V}$$ is the velocity of charged particle

Thus,from the expression of force it is clear that $$\vec{F}$$ is perpendicular to both $$\vec{V}and \vec{B}$$

As $$Force $$ is perpendicular to $$Velocity ,$$ this implies $$Speed $$ will not be changed only $$Direction $$ will be changed

Thus $$Speed $$ will remain unchanged

answer A
Question 7
1 / -0

A negative test charge is moving near a long straight current carrying wire . A force will act on the test charge in a direction parallel to the direction of the current, if the motion of the charge is in a direction

A
towards wire

B
the same as that of current

C
opposite to that of current

D
perpendicular to both the direction of current and the direction towards wire.

Solution

The magnetic force $$\vec{F}$$ shall act on the negatively charged test particle in direction parallel to current carrying wire, if the velocity $$\vec{V}$$ of the negative charge is as shown.
Question 8
1 / -0

A charged particle is kept at rest in a uniform magnetic field. If the magnetic field increases with time,

A
the particle starts moving

B
the particle undergoes circular motion

C
the particle starts moving in opposite direction

D
None of the above

Solution

A change in magnetic field will produce an electric field. Since the particle is charged, it will experience a force due to the electric field and will start moving.
Question 9
1 / -0

Two long thin conductors $$10$$ $$cm$$ apart carry currents in the ratio $$1 : 2$$ in the same direction. The magnetic field midway between them is $$2\times 10^{-3}$$ Tesla. The force on $$1$$ metre length of any conductor will be :

A
$$2\pi $$ x $$10^{-3}$$ Newton

B
$$2\times10^{-3}$$ Newton

C
$$\pi$$ Newton

D
$$1$$ Newton

Solution

Magnetic field at $$\displaystyle \mathrm{P}=\dfrac{\mu_{0}}{2\pi\bigg (\dfrac{\mathrm{d}}{2}\bigg)}[2i-i]=\dfrac{\mu_{0}\mathrm{i}}{\pi\mathrm{d}}$$
$$\cfrac{\mu _{o}i}{\pi d}=2\times 10^{-3}$$
$$i=\cfrac{2 \times 10^{-3} \times 0.1}{4 \times 10^{-7}}$$
or $$i=\displaystyle \cfrac{10^{3}}{2}$$
Magnetic field on II due to I = $$\cfrac{\mu _{o}i}{2\pi d}=B$$
Force on unit length of II $$= B \times 2i$$
$$=\dfrac{\mu _{o}i^{2}}{\pi d}$$
$$= 1$$ Newton
Question 10
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A particle of specific charge $$\alpha$$ is projected from origin with velocity $$\overrightarrow{v}=v_0\hat i-v_0\hat k$$ in a uniform magnetic field $$\overrightarrow{B}=-B_0\hat k$$. Find time dependence of velocity of the particle :

A
$$\overrightarrow{v}_{(t)}=v_0 \cos(\alpha B_0t)\hat i+v_0 \sin(\alpha B_0t)\hat j-v_0\hat k$$

B
$$\overrightarrow{v}_{(t)}=-v_0 \cos(\alpha B_0t)\hat i+v_0 \sin(\alpha B_0t)\hat j+v_0\hat k$$

C
$$\overrightarrow{v}_{(t)}=-v_0 \cos(\alpha B_0t)\hat i+v_0 \sin(\alpha B_0t)\hat j-v_0\hat k$$

D
$$\overrightarrow{v}_{(t)}=v_0 \cos(\alpha B_0t)\hat i+v_0 \sin(\alpha B_0t)\hat j+v_0\hat k$$

Solution