Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 65

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Question 1
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Two wires A & B are carrying currents $$I_1$$ & $$I_2$$ as shown in the figure. The separation between them is $$d$$. A third wire C carrying a current $$I$$ is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are :

A
$$x=\frac{I_1}{I_1-I_2}$$d and $$x=\frac{I_2}{I_1-I_2}d$$

B
$$x=\pm(\frac{I_1d}{I_1-I_2})$$

C
$$x=\frac{I_1}{I_1+I_2}$$d and $$x=\frac{I_2}{I_1-I_2}d$$

D
$$x=\frac{I_2}{I_1+I_2}$$d and $$x=\frac{I_1}{I_1-I_2}d$$

Solution

Net force on wire carrying current $$I$$ per unit length is
$$\dfrac{\mu_0 I_1I}{2\pi x}+\dfrac{\mu_0I_2I}{2\pi(d-x)}=0$$
$$\dfrac{I_1}{x}=\dfrac{I_2}{x-d}$$
$$\Rightarrow x=\dfrac{I_1d}{I_1-I_2}$$
Question 2
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The correct Biot-Savart law in vector form is?

A
$$d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{l}\times \vec{r})}{r^2}$$

B
$$d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{I(d\vec{l}\times \vec{r})}{r^3}$$

C
$$d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}}{r^2}$$

D
$$d\vec{B}=\dfrac{\mu_0}{4\pi}\cdot \dfrac{Id\vec{l}}{r^3}$$

Solution

Biot-savart law is given by
$$dB=\dfrac{\mu_0}{4\pi} \dfrac{idlsin\theta}{r^2}$$
vector form of biot-savart law is given by,
$$dB=\large { \dfrac{\mu_0}{4\pi}}\times \dfrac{i\overrightarrow{dl}\times \overrightarrow{r}}{\overrightarrow{r^3}}$$
SI unit B is tesla
magnetic field due to current carrying wire is given by,
$$B\rightarrow \int\dfrac{\mu_0}{4\pi}\dfrac{i(\overrightarrow{dl} \times \overrightarrow{r})}{r^3}$$
Question 3
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Which of the following gives the value of magnitude field according to, Biot-Savart's law'

A
$$ \frac {i\triangle l sin \theta}{r^2} $$

B
$$ \frac {\mu_o}{4 \pi} \frac {i \triangle l sin \theta}{r} $$

C
$$ \frac {\mu_o}{4\pi} \frac {i \triangle l sin \theta}{r^2} $$

D
$$ \frac {\mu_o}{4 \pi} i \triangle l sin \theta $$

Solution
Question 4
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An electron, a proton and a $$He^+$$ ion projected into a magnetic field with same kinetic energy, with velocities being perpendicular to the magnetic field. The order of the radii of cirlces traced by them is:

A
$$r_p > r_{He^+} > r_e$$

B
$$r_{He^+} > r_p > r_e$$

C
$$r_p = r_{He^+} > r_e$$

D
None of these

Solution

radius of circle is given by
$$r = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2 mk}}{qB} = \dfrac{\sqrt{2m}}{qB} \sqrt{k}$$
where K is kinetic energy
For poor
$$r_p = \dfrac{\sqrt{2m_p}}{eB} \sqrt{k}$$
for electron $$r_e = \dfrac{\sqrt{2m_e}}{eB} \sqrt{K}$$
for $$He^+ \, r_{He^+} = \dfrac{\sqrt{2 \times 4 m_p}}{eB} \sqrt{K} = \dfrac{\sqrt[2]{2m_p}}{eB} \sqrt{K}$$
Clearly $$r_{He^+} > r_p > r_e$$
Question 5
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A long straight wire carries a current of $$50\, A$$. An electron moving at $$10^7\, m/s$$ is $$5\, cm$$ away from the wire. The force acting on electron if its velocity is directed towards the wire will be :

A
$$1.6\times 10^{-6}\, N$$

B
$$3.2\times 10^{-16}\, N$$

C
$$4.8\times 10^{-16}\, N$$

D
$$1.8\times 10^{-16}\, N$$

Solution

Given: Current in wire= 50A
Speed of electron=$$10^{7}ms^{-1}$$
Solution: Let us assume the current is directed in +y direction.
As we know, magnetic field due to straight wire is given by,
$$B=\frac{\mu _{0}i}{2\pi r}$$
$$=\frac{4\pi \times 10^{-7}\times 50}{2\pi \times 0.05}$$
$$=0.0002T$$
According to question, v and B are perpendicular to each other.
$$F=qvb$$
Substituting the values we get,
$$=1.6\times 10^{-19}\times 10^{7}\times 0.0002$$
$$=3.2\times 10^{-16}N$$
Hence, the correct option is (B).
Question 6
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A stationary magnet does not intereact with

A
iron rod

B
moving charge

C
moving magnet

D
stationary charge

Solution

Solution
A static charge does not interact with a magnet.
Force will only applied by magnetic field on moving charge
The correct option is D
Question 7
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A particle of charge $$ 1 \mu C $$ is at rest in a magnetic field $$ \overrightarrow {B} = -2 \overrightarrow {k} $$ tesla,Magnetic Lorentz force on the charge particle with respect to an observer moving with velocity $$ \overrightarrow {v} = -5 \hat {i} ms^{-1} $$ will be

A
zero

B
$$ -10^{-5} \hat {j} N $$

C
$$ -10^{-6} \hat {j} N $$

D
$$ +10^{-5} \hat {j} N $$

Solution

Given
Charge= $$10^{-6}$$C
B= -2T $$ \hat{k}$$
v= -5m/s $$\hat{i}$$
Solution
Lorentz forve is reference dependent therefore for moving reference velocity of object is in opposite direction that of moving reference
Therefore velocity of charge= +5m/s $$\hat{i}$$
F=q(vxB)
$$F=10^{-6}(5*2)(\hat{i} * \hat{k})$$
F=10^{-5} $$\hat{j}$$
The correct option is D
Question 8
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A point charge $$q$$ is placed near a long straight wire carrying current $$I$$, then :

A
there is small charge density on the surface wire

B
there is no charge density on the surface wire

C
no force is exerted by wire on the point charge $$a$$

D
a large force is exerted by wire on the point charge $$a$$

Solution
Question 9
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If a long copper and carries a direct current, the magnetic field due to current will be :

A
inside the rod only

B
outside the rod only

C
both inside and outside the rod

D
neither inside nor outside the rod

Solution
Question 10
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In moving coil galvanometer, strong horses shoe magnet of concave shaped pole pieces is used to?

A
Increase space for rotation of coil

B
Reduce weight of galvanometer

C
Protect magnetic field which is parallel to plane of coil at any position

D
Make magnetic induction weak at the cnetre

Solution

The moving coil of a moving coil galvanometer moves in a magnetic field produced by a permanent magnet. When a current passes through the coil, its sides which are perpendicular to the magnetic field, experience equal and opposite force. This force is separated by the width of the coil. These two equal and opposite forces separated by the width of the coil constitute a couple, which rotates the coil.
We want the current-carrying coil to be always perpendicular to the magnetic field, even when it has rotated. The magnetic field produced by concave-shaped pole faces is always radial. A radial field is always perpendicular to a conductor rotating about an axis passing through the centre of the concave-shaped pole faces and parallel to the faces.
Pole faces are concave to make the magnetic field radial to keep it always normal to the moving coil.