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CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 6 - MCQExams.com
CBSE
Class 12 Medical Physics
Moving Charges And Magnetism
Quiz 6
A proton is moving with a velocity of $$3\times 10^7 m/s$$ in the direction of a uniform magnetic field of $$0.5 T$$. The force acting on proton is
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$$2 N$$
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$$4 N$$
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$$6 N$$
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zero
Explanation
$$\overrightarrow{F}_{magnetic} = q(\overrightarrow{v} \times \overrightarrow{B})$$
Since $$\overrightarrow{v} \parallel \overrightarrow{B}$$
$$\overrightarrow{F}_{magnetic} =0$$
A current of 30 amp is flowing in a conductor as shown in the figure. The magnetic induction at point O will be :
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$$1.5 \ Tesla$$
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$$4.71\times 10^{-4}Tesla$$
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zero
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$$0.15\ Tesla$$
Explanation
Magnetic field produced by a circular coil at its center is $$B= \dfrac{\mu_0i}{2R}$$
Since the length of the given portion of the coil is $$\dfrac{3}{4}$$ of the perimeter of the circular coil, the magnetic field produced by it will be $$\dfrac{3}{4}$$th of that produced by the circular coil.
$$ \therefore B = \dfrac{3}{4} \times \dfrac{\mu_0i}{2R}=\dfrac{3\mu_0i}{8R} = \dfrac{ 3 \times 4\pi \times 10^{-7} \times 30 }{8 \times 3 \times 10 ^{-2}}= 4.71 \times 10^{-4}\: Tesla$$
A charge of $$0.04 C$$ is moving in a magnetic field of $$0.02 T$$ with a velocity $$10 m/s$$ in a direction making an angle $$30^o$$ with the direction of field. The force acting on it will be :
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$$4\times 10^{-3} N$$
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$$2\times 10^{-3} N$$
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zero
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$$8\times 10^{-3} N$$
Explanation
Force on the moving charge is,
$$\vec F=q(\vec{v} \times \vec{B})= qvB \sin \theta= 0.04 \times 10 \times 0.02 \sin 30^o= 0.04 \times 10 \times 0.02 \times 0.5 = 4 \times 10^{-3} \:N$$
A wire is lying parallel to a square coil. Same current is flowing in same direction in both of them. The magnetic induction at any point P inside the coil will be :
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zero
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more than that produced by only coil
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less than that produced by only coil
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equal to that produced by only coil
Explanation
$$\vec{B_{QR}} = \vec{B_{SP}}$$ and are $$-\hat{k}$$ direction.
$$\vec{B_{PQ}}$$ and $$\vec{B_{RS}} $$ are in $$-\hat{k}$$ direction.
Magnetic field due to the long wire is also in $$-\hat{k}$$ direction.
Since all are in the same direction, magnetic field at point P( center of the coil) is more than that produced by the coil.
On passing electric current in two long straight conductors in mutually opposite directions, the magnetic force acting between them will be
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attractive
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repulsive
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both attractive and repulsive
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neither attractive nor repulsive
Explanation
If electric current in two long straight conductors is in mutually opposite directions, the magnetic force acting between them will be
repulsive.
A charged particle of mass $$10^{-3}$$ kg and charge $$10^{-5}C$$ enters a magnetic field of induction 1 T. If $$g=10 ms^{-2}$$ for what value of velocity will it pass straight through the field without deflection?
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$$10^{-3} ms^{-1}$$
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$$10^{3} ms^{-1}$$
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$$10^{6} ms^{-1}$$
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$$1 ms^{-1}$$
Explanation
Since the charged particle passes without being deflected,
$$F_{magnetic} = F_{gravity}$$
$$ \therefore qvB = mg$$
$$ \therefore v=\dfrac{mg}{qB}$$
$$=\dfrac{10^{-3} \times 10}{10 ^{-5} \times 1}=10^3\: m/s$$
Two current-carrying parallel conductors are shown in the figure. The magnitude and nature of force acting between them per unit length will be :
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$$8\times 10^{-8}N/m$$, attractive
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$$3.2\times 10^{-5}N/m$$, repulsive
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$$3.2\times 10^{-5}N/m$$, attractive
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$$8\times 10^{-8}N/m$$, repulsive
Explanation
Force on a current carrying conductor is $$F=\int i (\vec{dl} \times \vec{B})$$
Magnetic field due to one wire at the location of the second wire ( other one ) is $$B_1= \dfrac{\mu_0 i_1}{2\pi d}$$
Force on the 2nd wire due to $$B_1$$ is $$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$$
Force per unit length for wire $$l_2$$ is $$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}=\dfrac{\mu_0 i_1i_2}{2\pi d}=\dfrac{2 \times 10^{-7} \times 4 \times 4 }{10 \times 10^{-2}}=3.2 \times 10^{-5}\: N$$
where $$F_{21}$$ is the force on wire 2 due to wire 1 and the subscript 2 refers to 2nd wire.
$$B_1$$ is in $$-\hat{k}$$ direction.
$$l_2$$ is in $$-\hat{j}$$ direction.
Hence, $$F_{21}$$ will be in $$\hat{j} \times (-\hat{k})=-\hat{i}$$ direction.
Similarly, $$F_{12}$$ will be in $$\hat{i}$$ direction.
Thus, both the wires will attract each other.
A long wire is bent into the shape $$PQRST$$ as shown in the following Figure with $$QRS$$ being a semicircle with centre $$O$$ and radius r metre. A current of $$I$$ ampere flows through it in the direction $$P\rightarrow Q\rightarrow R\rightarrow S\rightarrow T$$. Then, the magnetic induction at the point $$O$$ of the figure in vacuum is
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$$\mu_0i\left [\dfrac {1}{2\pi r}+\dfrac {1}{4r}\right ]$$
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$$\mu_0i\left [\dfrac {1}{2\pi r}-\dfrac {1}{4r}\right ]$$
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$$\dfrac {\mu_0i}{4r}$$
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$$\dfrac {\mu_0i}{\pi r}$$
Explanation
Magnetic field at $$O$$ due to the semi-infinite segment $$PQ$$ is half of that due to an infinite wire.
$$ \vec{B_{PQ}}= \dfrac{1}{2}\dfrac{\mu_0 i}{2\pi r} (-\hat{k})$$
Similarly,
$$ \vec{B_{ST}}= \dfrac{1}{2}\dfrac{\mu_0 i}{2\pi r} (-\hat{k})$$
$$ \vec{B_{QRS}} = \dfrac{1}{2} \dfrac{\mu_0 i}{4r}$$ where $$\frac{\mu_0 i}{2r} (\hat{k}$$ is the magnetic field due to a circular wire at the center of the circle.
The magnetic field due to semi circular portion is half of that due to a circular wire.
$$ \therefore \vec{B_{total}} = \vec{B_{PQ}} + \vec{B_{ST}} + \vec{B_{QRS}} = \mu_0 i\left[ \dfrac{1}{2\pi r} + \dfrac{1}{4r} \right]$$
Current of 10 ampere and 2 ampere are passed through two parallel wires A and B, respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is 2 m, the force on the conductor B which is situated at 10 cm distance from A will be
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$$8\times 10^{-5}N$$
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$$4\times 10^{-5}N$$
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$$8\pi \times 10^{-7}N$$
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$$4\pi \times 10^{-7}N$$
Explanation
Using Amper's law $$\oint \vec{B}.\vec{dl} = \mu_0 i_{enclosed}$$
$$ \therefore B = \dfrac{\mu_0 i_{enclosed}}{2\pi r}= \dfrac{4\pi \times 10^{-7} \times 10 }{2 \pi \times(10 \times 10 ^{-2})}= 2 \times 10^{-5} \:T$$
Force on the conductor $$B$$ is $$F= ilB = 2 \times 2 \times (2 \times 10 ^{-5})= 8 \times 10^{-5} \:N$$
Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. the force of attraction per unit length between the two wires is
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$$2\times 10^{-7}Nm^{-1}$$
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$$2\times 10^{-8}Nm^{-1}$$
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$$5\times 10^{-8}Nm^{-1}$$
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$$10^{-7}Nm^{-1}$$
Explanation
Force on a current carrying conductor is $$F=\int i (\vec{dl} \times \vec{B})$$
Magnetic field due to one wire at the location of the second wire ( other one ) is $$B_1= \dfrac{\mu_0 i_1}{2\pi d}$$
Force on the 2nd wire due to $$B_1$$ is $$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$$
Force per unit length for wire $$l_2$$ is $$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}=\frac{\mu_0 i_1i_2}{2\pi d}=\dfrac{2 \times 10^{-7} \times 1 \times 1 }{1}=2 \times 10^{-7} \: N$$
where $$F_{21}$$ is the force on wire 2 due to wire 1 and the subscript 2 refers to 2nd wire.
$$B_1$$ is in $$-\hat{k}$$ direction.
$$l_2$$ is in $$-\hat{j}$$ direction.
Hence, $$F_{21}$$ will be in $$\hat{j} \times (-\hat{k})=-\hat{i}$$ direction.
Similarly, $$F_{12}$$ will be in $$\hat{i}$$ direction.
Thus, both the wires will attract each other.
The initial acceleration of the proton is
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$$2.9\times 10^8 ms^{-2}$$
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$$3.31\times 10^2 ms^{-2}$$
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$$3.12\times 10^8 ms^{-2}$$
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none of these
Explanation
Current $$I_1=\dfrac{100}{10}A$$ = $$10A$$
Current $$I_2=\dfrac{100}{25}A$$ = $$4A$$
Force on moving proton due to magnetic field is $$F=qvB$$
$$F=qv\left[ \cfrac {\mu_0I_1}{2\pi d}-\cfrac {\mu_0I_2}{2\pi d}\right]=\cfrac{qv\mu_0}{2\pi d} \left[I_1- I_2 \right]$$
$$ma=\cfrac{qv\mu_0}{2\pi d} \left[I_1-I_2 \right]$$
$$a=\dfrac {1.6\times 10^{-19}\times 650\times 10^3[10-4]\times 4\pi \times 10^{-7}}{2\pi \times 2.5\times 10^{-2}\times 1.6\times
10^{-27}}$$
$$a =3.12\times 10^8ms^{-2}$$
The radius of the curved part of the wire is $$R$$, the linear parts are assumed to be very long.
Find the magnetic induction of the field at the point $$O$$ if a current-carrying wire has the shape shown in figure above.
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$$\displaystyle B=\dfrac{\mu_0}{4\pi}\dfrac{I}{R}\left[1+\dfrac{3\pi}{2}\right]$$
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$$\displaystyle B=\dfrac{\mu_0}{\pi}\dfrac{I}{R}\left[1+\dfrac{3\pi}{2}\right]$$
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$$\displaystyle B=\dfrac{\mu_0}{2\pi}\dfrac{I}{R}\left[1+\dfrac{3\pi}{2}\right]$$
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$$\displaystyle B=0$$
Explanation
Here, we consider magnetic induction due to long linear part,
$$B_1 = \dfrac{\mu_0 I}{4 \pi R}(\cos 0^o - \cos 90^o) = \dfrac{\mu_0 I}{4 \pi R}$$
Now, the magnetic induction $$dB_2$$ due to small curved part $$dl$$ is
$$dB_2 = \dfrac{\mu_0 I}{4\pi R^2} \left(\dfrac{r\sin 90^o}{r^3}dl \right)$$
$$B_2 = \dfrac{\mu_0 I}{4\pi R^2}\left(\dfrac{3}{4}\right)2\pi R$$
$$B_2 = \dfrac{3\pi\mu_0 I}{8\pi R}$$
The magnetic induction of the field at the point $$O$$
is
$$B = B_1 + B_2$$
$$B = \dfrac{\mu_0 I}{4 \pi R} +\dfrac{3\pi\mu_0 I}{8\pi R}$$
$$B = \dfrac{\mu_0 I}{4 \pi R}\left[1 + \dfrac{3\pi}{2}\right]$$
The radius of the curved part of the wire is $$R$$, the linear parts are assumed to be very long.
Find the magnetic induction of the field at the point $$O$$ if a current-carrying wire has the shape shown in figure above.
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$$\displaystyle B=\dfrac{3\mu_0}{4}\dfrac{i}{R}$$
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$$\displaystyle B=\dfrac{\mu_0}{2}\dfrac{i}{R}$$
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$$\displaystyle B=\dfrac{2\mu_0}{3}\dfrac{i}{R}$$
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$$\displaystyle B=\dfrac{\mu_0}{4}\dfrac{i}{R}$$
Explanation
Magnetic field at point O due to part (1) & (3)$$= 0$$ (magnetic field along a straight line wire is always zero)
Now magnetic field at point O due to part (2) $$\Rightarrow {\dfrac {1}{2}}{\times}{\dfrac {{{\mu}_{0}}{i}}{2R}} $$
$$\Rightarrow{\dfrac {{{\mu}_{0}}{i}}{4R}} $$
The work done by a magnetic field, on a moving charge is
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zero because $$\vec F$$ acts parallel to $$\vec v$$
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positive because $$\vec F$$ acts perpendicular to $$\vec v$$
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zero because $$\vec F$$ acts perpendicular to $$\vec v$$
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negative because $$\vec F$$ acts parallel to $$\vec v$$
Explanation
Force on moving charge while moving in magnetic field is;
$$\vec = q (\vec v \times \vec B)$$
where $$\vec F$$ is perpendicular to $$\vec v$$.
Work done/sec $$= \vec F . \vec v = F \vec v cos 90^o = 0$$
The radius of the curved part of the wire is $$R$$, the linear parts are assumed to be very long.
Find the magnetic induction of the field at the point $$O$$ if a current-carrying wire has the shape shown in figure above.
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$$\displaystyle B=\dfrac{\mu_0}{\pi}\dfrac{i}{R}(2+\pi)$$
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$$\displaystyle B=\dfrac{\mu_0}{4\pi}\dfrac{i}{R}(4+\pi)$$
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$$\displaystyle B=\dfrac{\mu_0}{2\pi}\dfrac{i}{R}(4+\pi)$$
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$$\displaystyle B=\dfrac{\mu_0}{4\pi}\dfrac{i}{R}(1+\pi)$$
Explanation
Here, we consider magnetic induction due to long linear part,
$$B_1 = \dfrac{\mu_0 I}{4 \pi R}(\cos 0^o - \cos 90^o) = \dfrac{\mu_0 I}{4 \pi R}$$
Since there are 2 linear parts, the magnetic field due to them = $$2B_1$$
Now,
magnetic induction $$dB_2$$ due to small curved part $$dl$$ is
$$dB_2 = \dfrac{\mu_0 I}{4\pi R^2} \left(\dfrac{r\sin 90^o}{r^3}dl \right)$$
$$B_2 = \dfrac{\mu_0 I}{4\pi R^2}\left(\dfrac{1}{2} \right)2\pi R$$
$$B_2 = \dfrac{\pi\mu_0 I}{4\pi R}$$
The magnetic induction of the field at the point $$O$$ is
$$B = 2B_1 + B_2$$
$$B = 2\dfrac{\mu_0 I}{4 \pi R} + \dfrac{\pi\mu_0 I}{4\pi R}$$
$$B = \dfrac{\mu_0 I}{4 \pi R} \left[2 + \pi \right] $$
Find the magnetic moment of the spiral with a given current.
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$$\displaystyle p=25\:mA\cdot m^2$$
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$$\displaystyle p=15\:mA\cdot m^2$$
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$$\displaystyle p=30\:mA\cdot m^2$$
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$$\displaystyle p=50\:mA\cdot m^2$$
Explanation
$$\displaystyle p=\frac{\pi iN(b^3-a^2)}{3(b-a)}=15\:mA\cdot m^2$$ (we have used the standard formula for the required calculation.)
An electron accelerated through a potential difference V passes through a uniform transverse magnetic field and experiences a force F. If the accelerating potential is increased to 2V, the electron in the same magnetic field will experience a force
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F
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F/2
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$$\sqrt 2F$$
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2F
Explanation
$$F=Bqv$$
But $$\frac {1}{2}mv^2=eV$$ or $$v=\sqrt {\frac {2eV}{m}}$$
$$\therefore F=Bq\sqrt {\frac {2eV}{m}}$$
$$\Rightarrow F\propto \sqrt V$$ and
$$F\propto \sqrt {2V}$$
$$\frac {F'}{F}=\sqrt 2$$ or $$F'=\sqrt 2F$$
An electron and a proton are injected into a uniform magnetic field perpendicular to it with the same momentum. If both particles are fired with same momentum into a transverse electric field, then
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electron trajectory is less curved
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proton trajectory is less curved
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both trajctories are equally curved
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both trajectories are straight lines
Explanation
let, momentum of electron and proton be p.
Transverse electric force on particle provides the centripetal force to support the circular motion.
If r is radius of curvature, $$ F_e = F_c $$
i.e. $$qE = m \dfrac{v^2}{r} $$.
Solving the above equation we get r = $$ \dfrac {p^2}{mqE} $$
where, $$p = momentum = mv$$
Given $$q$$ ,$$p$$ and $$E$$ are constant. So, we can say that $$ r \propto \dfrac{1}{m} $$
Radius of curvature is higher for a lighter mass particle which is an electron and vice versa for proton.
A horizontal circular loop carries a current that looks clockwise when viewed from above. It is placed by an equivalent magnetic dipole consisting of a south pole $$S$$ and a north pole $$N$$.
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The line $$SN$$ should be along a diameter of the loop.
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The line $$SN$$ should be perpendicular to the plane of the loop.
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The south pole should be below the loop.
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The north pole should be below the loop.
Explanation
Since a current carrying circular loop behaves as a magnetic dipole whose magnetic moment's direction is perpendicular to the plane of the circular loop.
therefore, when this dipole is replaced by equivalent bar magnet then the line joining NS will be perpendicular to the plane of the loop
Now the direction of current flowing through the loop is clockwise when viewed from the top.
Therefore, magnetic field at the center of the loop is inwards therefore, South pole of the magnet should be above the plane and North pole of the magnet should be below the plane.
A conducting rod PQ is moving parallel to X-axis in a uniform magnetic filed directed in positive Y-direction. The end P of the rod will become
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negative
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positive
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neutral
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sometimes negative
Explanation
Force of positive charges in rod = $$ q \vec v \times \vec B $$
As $$v$$ is along x axis and $$B$$ is along y axis the force will be along the z axis. Therefore the positive charges will move towards the end P and it will become positive charged.
An electron and a proton are injected into a uniform magnetic field perpendicular to it with the same momentum. If the two particles are injected into a uniform transverse electric field with same kinetic energy, then
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electron trajectory is more curved
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proton trajectory is more curved
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both trajectories are equally curved
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both trajectories are straight lines
Explanation
Let, Kinetic energies of electron and proton be K.
Transverse electric force on particle provides the centripetal force to support the circular motion in this case.
If r is radius of curvature, $$ F_e = F_c $$
i.e. qE = m $$\dfrac{v^2}{r} $$.
Solving the above equation we get r = $$ \dfrac {2K}{qE} $$ where K = Kinetic Energy = $$\dfrac {mv^2}{2}$$
Here, q,E and K are constant. So, radius of curvature is same for both of them.
A person is facing magnetic north. An electron in front of him flies horizontally towards the north and deflects towards east. He is in/at the
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southern hemispheres.
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the equator
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northern hemispheres.
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none of these
Explanation
Now electron moves towards north and get deflected towards east therfore, vertical component of earth's magnetic field must be positive. Now when we find the direction of the force acting on electron by finding the direction of vector $$\overrightarrow { v } \times \overrightarrow { { B }_{ v } } $$
(We will neglect the horizontal component because it will be parallel to the direction of motion of electron hence $$\overrightarrow { v } \times \overrightarrow { { B }_{ H } }=0 $$).
T
he direction of this vector is towards west when $$B_v$$ is pointing downwards and our charge is electron therefore, force on it will exert in the direction of east.
Now vertical component is downward in southern hemisphere.
A current $$i$$ is flowing in a conductor of length $$l$$. When it is bent in the form of a loop its magnetic moment will be
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$$\displaystyle 4\pi l^2 i$$
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$$\dfrac{il^2}{4\pi}$$
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$$\dfrac{4\pi}{l^2i}$$
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$$\dfrac{l^2}{4\pi}$$
Explanation
When the wire of length $$l$$ is bent to form a loop of radius $$r$$
$$r=\dfrac{l}{2\pi}$$
Area of the loop $$A=\pi r^2 = \pi\left(\dfrac{l}{2\pi}\right)^2$$
Magnetic moment: $$\mu = iA = i\pi \dfrac{ l^2}{4\pi ^2}=\dfrac{il^2}{4\pi}$$
The resultant magnetic moment due to two currents carrying concentric coils of radius $$r$$, mutually perpendicular to each other will be
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$$\sqrt{2}ir$$
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$$\sqrt{2}i \pi r^2$$
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$$2\pi r^2$$
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$$\sqrt{2}ir^2$$
Explanation
Magnetic moment of coil-1 $$\mu_1 = i\pi r^2$$
Magnetic moment of coil-2 $$\mu_2 = i\pi r^2$$
$$\mu_1 \perp \mu_2$$
$$\therefore$$ resultant of $$\mu_1$$ and $$\mu_2$$ is $$\mu = \sqrt{(\mu_1^2 + \mu_2^2)}=\sqrt{2}\mu_1=\sqrt{2} i\pi r^2$$
When a conductor is rotated in a perpendicular magnetic field then, it's free electrons
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move in the field direction.
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move at right angles to field direction.
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remain stationary.
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move opposite to field direction.
Explanation
Force on electron = $$ -q_e \vec v \times \vec B $$
the force is perpendicular to the magnetic field.
Two thin, long, parallel wires, separated by a distance 'd' carry a current of 'i' A in the same direction. They will
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repel each other with a force of $$\mu_0 i^2/(2 \pi d)$$
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attract each other with a force of $$\mu_0 i^2 / (2\pi d)$$
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repel each other with a force of $$\mu_0i^2 / (2 \pi rd^2)$$
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attract each other with a force of $$\mu_0 i^2 / (2 \pi d^2)$$
Explanation
$$\displaystyle \dfrac{F}{l} = \dfrac{\mu_0 i_1 i_2}{2 \pi d} $$
$$= \dfrac{\mu_0 i^2}{2 \pi d}$$
(attractive as current is in the same direction)
The magnetic filed (dB) due to smaller element (dl) at a distance $$(\vec r)$$ from element carrying current i, is
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$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} \left ( \frac{\vec{dl} \times \vec r}{r} \right )$$
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$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} i^2 \left ( \frac{\vec{dl} \times \vec r}{r^2} \right )$$
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$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} i^3 \left ( \frac{\vec{dl} \times \vec r}{2r^2} \right )$$
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$$\displaystyle dB = \frac{\mu_0}{4 \pi} i \left ( \frac{\vec{dl} \times \vec r}{r^3} \right )$$
Explanation
$$dB=\dfrac { { \mu }_{ 0 }i }{ 4\pi } \int { \dfrac { \left( \overrightarrow { dl } \times \hat { r } \right) }{ { r }^{ 2 } } } \\$$
we know that=$$\hat { r } =\dfrac { \overrightarrow { r } }{ { r }} \\$$
$$dB=\dfrac { { \mu }_{ 0 }i }{ { 4 }\pi } \int { \dfrac { \left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ { r }^{ 3 } } }$$
An electron is accelerated from rest through a potential difference $$V$$. This electron experiences of force $$F$$ in a uniform magnetic field. On increasing the potential difference to $$V^{'}$$, the force experienced by the electron in the same magnetic field becomes $$2F$$. Then, the ratio $$\dfrac{V^{'}}{V}$$ is equal to
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$$\dfrac{4}{1}$$
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$$\dfrac{2}{1}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{4}$$
Explanation
$$K.E. = \dfrac{1}{2}mv^2 = eV$$
$$ \therefore v =\sqrt{\dfrac{2eV}{m}}$$
$$ F =eVB =e\sqrt{\dfrac{2eV}{m}}B$$
Given $$F' =2F = e\sqrt{\dfrac{2eV^{'}}{m}}B$$
$$\Rightarrow 2e\sqrt{\dfrac{2eV}{m}} B = e\sqrt{\dfrac{2eV'}{m}}B$$
$$ \Rightarrow V^{'} = 4V$$
A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one end. If a magnetic field is switched on in the vertical direction, the tension in the string
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will increase.
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will decrease.
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remains same.
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may increase or decrease.
Explanation
Let the string makes an angle $$\theta$$ with the vertical.
Initially, before the magnetic field is applied, $$T \sin \theta= \dfrac{mv^2}{r}$$
If the particle experiences a force outward in the direction of the radius, tension $$T$$ will increase.
If the particle experiences a force inward in the direction of the radius towards the center, tension $$T$$ will decrease.
The force $$\vec F$$ experienced by a particle of charge q moving with a velocity $$\vec v$$ in a magnetic field $$\vec B$$ is given by $$\vec F=q(\overrightarrow{v}\times \overrightarrow{B})$$. Which pairs of vectors are always at right angles to each other?
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$$\overrightarrow{F}$$ and $$\overrightarrow{v}$$
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$$\overrightarrow{F}$$ and $$\overrightarrow{B}$$
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$$\overrightarrow{B}$$ and $$\overrightarrow{v}$$
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$$\overrightarrow{F}$$ and $$\left(\overrightarrow{v}\times \overrightarrow{B}\right)$$
Explanation
The pairs $$\overrightarrow{F}, \overrightarrow{v}$$ and $$\overrightarrow{F}, \overrightarrow{B}$$ are always at right angle to each other, because $$\overrightarrow{F}$$ is always perpendicular to the plane containing $$\overrightarrow{B}$$ and $$\overrightarrow{v}$$. Vectors $$\overrightarrow{B}$$ and $$\overrightarrow{v}$$ may have any angle between them.
$$\vec{F} =q(\vec{v} \times \vec{B})$$
From the defn of vector product, $$\vec{F} \perp \vec{v}$$ and $$\vec{F} \perp \vec{B}$$
A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be
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a circle
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a helix with uniform pitch
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a helix with non-uniform pitch
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a helix with uniform radius
Explanation
A charged particle moves in a circle when its velocity is perpendicular to the magnetic field. When it forms an acute angle with the magnetic field, it can be resolved in two components, parallel and perpendicular. The perpendicular components tends to move it in circle, the parallel components tends it to move along the magnetic field to form a helical motion of uniform radius and pitch.
A charged particle moves with velocity $$\vec v=a\hat i+d\hat j$$ in a magnetic field $$\vec B=A\hat i+D\hat j$$. The force acting on the particle has magnitude $$F$$. Then,
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$$F=0$$, if $$aD=dA$$
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$$F=0$$, if $$aD=-dA$$
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$$F=0$$, if $$aA=-dD$$
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$$F\propto (a^2+b^2)^{1/2}\times (A^2+D^2)^{1/2}$$
Explanation
$$\vec{F} = q(\vec{v} \times \vec{B})$$
$$=q((a\hat{i} + d\hat{j} ) \times ( A\hat{i} + D\hat{j}))$$
$$=q( aD-dA)\hat{k}$$
If $$aD-dA=0$$
$$ \vec{F} =0$$
A proton is fired from origin with velocity $$\overrightarrow{v}=v_0\hat {j}+v_0\hat{k}$$ in a uniform magnetic field $$\overrightarrow{B}=B_0\hat{j}$$. In the subsequent motion of the proton
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its $$y-$$coordinate will be proportional to its time of flight
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its $$x-$$coordinate can never by positive
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its $$x-$$ and $$z-$$coordinate cannot be zero at the same time
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None of the above
Explanation
$$\overrightarrow{F} =q(\overrightarrow{v} \times \overrightarrow{B})=q((v_o\hat{j} +v_o\hat{k} ) \times B_0\hat{j})=-v_0B_0\hat{i}$$
The force will be always along -ve x-axis.
$$\therefore$$, the particle will never have $$x-$$cordinate positive since it starts from the origin.
The velocity component along $$y-$$axis will make the path helical.
$$Y$$ co-ordinate will be $$y=v_0t$$ i.e $$y \propto t$$
A charged particle is fired at an angle $$\theta$$ to a uniform magnetic field directed along the x-axis. During its motion along a helical path, the particle will
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never move parallel to the $$x-$$axis
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move parallel to the $$x-$$axis once during every rotation for all values of $$\theta$$
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move parallel to the $$x-$$axis at least once during every rotation if $$\theta=45^o$$
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never move perpendicular to the $$x-$$direction
Explanation
The component of velocity $$v$$ along the direction of $$B$$ is $$v\cos \theta$$.
Since the trajectory will be helical, the trajectory will never be perpendicular to the direction of magnetic field $$B$$, it's trajectory will be inclined at a small angle to the direction of the magnetic field.
It will also never move parallel to $$x-$$axis.
A current of $$\dfrac{1}{(4\pi)}$$ ampere is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current-carrying conductor is
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$${10}^{-7}Wb$$ $${m}^{-1}$$
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$$4\pi \times{10}^{-7}Wb$$ $${m}^{-1}$$
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$$16{\pi}^{2} {10}^{-7}Wb$$ $${m}^{-1}$$
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$$zero$$
Explanation
$$\oint { \vec { B } } \cdot d\vec { l } ={ \mu }_{ 0 }I$$
$$=4\pi \times {
10 }^{ -7 }\times \cfrac { 1 }{ 4\pi } $$
$$= { 10 }^{ -7 }Wb{ m }^{
-1 }$$
The magnetic field at center $$O$$ of the arc in figure is
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$$\cfrac { { \mu }_{ 0 }I }{ 4\pi \times r } \left[ \sqrt { 2 } +\pi \right] $$
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$$\cfrac { { \mu }I }{ 2\pi r } \left[ \cfrac { \pi }{ 4 } +\left( \sqrt { 2 } -1 \right) \right] $$
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$$\cfrac { { { \mu }_{ 0 } } }{ 4\pi } \times \cfrac { I }{ r } \left[ \left( \sqrt { 2 } -\pi \right) \right] $$
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$$\cfrac { { { \mu }_{ 0 } } }{ 4\pi } \times \cfrac { I }{ r } \left[ \left( \sqrt { 2 } +\cfrac { \pi }{ 4 } \right) \right] $$
Explanation
Magnetic field due to straight wires
$${ B }_{ circular }=\cfrac { 1 }{ 4 } =\cfrac { { \mu }_{ 0 }I }{ 2r } =\cfrac { { \mu }_{ 0 }I }{ 8r } $$
$${ B }_{ straight }=2\cfrac { { \mu }_{ 0 }I }{ 4\pi (r\cos { { 45 }^{ o } } ) } \left[ \sin { { 90 }^{ o } } -\sin { { 45 }^{ o } } \right] $$
$$=\cfrac { { \sqrt { 2 } \mu }_{ 0 }I }{ 2\pi r } \left[ 1-\cfrac { 1 }{ \sqrt { 2 } } \right] $$
Currents $${I}_{1}$$ and $${I}_{2}$$ flow in the wires shown in figure. The field is zero at distance $$x$$ to the right of $$O$$. Then
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$$x=\left( \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } \right) a$$
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$$x=\left( \cfrac { { I }_{ 2 } }{ { I }_{ 1 } } \right) a$$
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$$x=\left( \cfrac { { I }_{ 1 }-{ I }_{ 2 } }{ { { I }_{ 1 }+I }_{ 2 } } \right) a$$
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$$x=\left( \cfrac { { I }_{ 1 }+{ I }_{ 2 } }{ { { I }_{ 1 }-I }_{ 2 } } \right) a$$
Explanation
Magnetic field with both wire will be equal and opposite in diretion,
$$\cfrac { { \mu }_{ 0 }{ I }_{ 1 } }{ 4\pi (a+x) } =\cfrac { { \mu }_{ 0 }{ I }_{ 2 } }{ 4\pi (a-x) } $$
$$\cfrac{ a-x }{ a+x } =\cfrac { { I }_{ 2 } }{ { I }_{ 1 } }$$
$$ \Rightarrow
x=\left( \cfrac { { I }_{ 1 }-{ I }_{ 2 } }{ { I }_{ 1 }+{ I }_{ 2 } }
\right) a$$
Hence Option C
The force on the charged particle in magnitude is
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$$23.04\times {10}^{-6}N$$
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$$230.4\times {10}^{-5}N$$
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$$0$$
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None of these
Explanation
Force acting on a charged particle moving in a magnetic field is
$$\vec{F}=q(\vec{v}\times \vec{B})$$
$$\implies \left|\vec{F}\right|=qvB$$
$$=1.6\times 10^{-19}\times 4.8\times 10^6\times 30\times 10^6N$$
$$=23.04\times 10^{-6}N$$.
Two long thin wires $$ABC$$ and $$DEF$$ are arranged as shown in figure. They carry equal current $$I$$ as shown. The magnitude of the magnetic field at $$O$$ is
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zero
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$${\mu}_{0}I/4\pi a$$
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$${\mu}_{0}I/2\pi a$$
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$${\mu}_{0}I/2 \sqrt {2}\pi a$$
Explanation
$$AB$$ and $$DE$$ do not give any magnetic field at $$O$$.
Magnetic field at a distance $$d$$ due to semi infinite wire is $$B = \dfrac{\mu_0 I}{4\pi d}$$
Magnetic field at $$O$$ due to $$BC$$ is $$B_{BC}=\cfrac { { \mu }_{ 0 }I }{ 4\pi a } $$ (out of plane of paper)
Magnetic field at $$O$$ due to $$EF$$ is
$$B_{EF}=\cfrac { { \mu }_{ 0 }I }{ 4\pi a } $$ (out of plane of paper)
So the final $$B$$ is the sum of the BC and EF.
So, $$B_{total} = B_{BC}+B_{EF}=\cfrac { { \mu }_{ 0 }I }{ 2\pi a } $$
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0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
Explanation
$$\vec{F} =q(\vec{v} \times \vec{B})$$
$$ \Rightarrow \vec{F} \perp \vec{v}$$
$$ \Rightarrow \vec{F}.\vec{v} = 0$$ since both are perpendicular to each other
$$ \Rightarrow KE =constant$$
The magnetic field at $$O$$ due to current in the wire segment BC of the infinite wire forming a loop as shown in figure is
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$$\cfrac { { { \mu }_{ 0 }I } }{ 4\pi d } \left( \cos { { \phi }_{ 1 } } +\cos { { \phi }_{ 2 } } \right) $$
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$$\cfrac { { { \mu }_{ 0 } } }{ 4\pi } \cfrac { 2I }{ d } $$
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$$\cfrac { { { \mu }_{ 0 } } }{ 4\pi } \cfrac { I }{ d } \left( \sin { { \phi }_{ 1 } } +\sin { { \phi }_{ 2 } } \right) $$
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$$\cfrac { { { \mu }_{ 0 } } }{ 4\pi } \cfrac { I }{ d } \ $$
Explanation
Using $$B=\cfrac { { \mu }_{ 0 }I }{ 4\pi a } \left[ \sin { { \theta }_{ 1 } } +\sin { { \theta }_{ 2 } } \right] $$
But $$\quad { \theta }_{ 1 }+{ \phi }_{ 1 }={ 90 }^{ o }\quad or\quad { \theta }_{ 1 }=\quad { 90 }^{ o }-\quad { \phi }_{ 1 }$$
$${ \phi }_{ 1 }=\sin { \left( { 90 }^{ o }-\quad { \phi }_{ 1 } \right) } \quad =\cos { { \phi }_{ 1 } } $$
Similarly
$$\sin { { \theta }_{ 2 } } =\cos { { \phi }_{ 2 } } $$
$${ B }_{ net }=\cfrac { { \mu }_{ 0 }I }{ 4\pi d } \left( \cos { { \phi }_{ 1 } } +\cos { { \phi }_{ 2 } } \right) $$
Three long, straight and parallel wires are arranged as shown in figure. The force experienced by $$10cm$$ length of wire $$Q$$ is
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$$1.4\times {10}^{-4}N$$ toward the right
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$$1.4\times {10}^{-4}N$$ toward the left
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$$2.6\times {10}^{-4}N$$ toward the right
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$$2.6\times {10}^{-4}N$$ toward the left
Explanation
Magnetic field produced by wire R at Q: $$B_1 = \dfrac{\mu_0i_1}{2\pi r_1} = \dfrac{\mu_0 \times 20}{2\pi \times ( 2 \times 10^{-2})}$$
Magnetic field produced by wire P at Q: $$B_2 = \dfrac{\mu_0i_2}{2\pi
r_2} = \dfrac{\mu_0 \times 30}{2\pi \times ( 10 \times 10^{-2})}$$
$$B_{net} =B_1 -B_2= \dfrac{\mu_0 \times 7}{2\pi \times 10^{-2}}$$
Force on 10 cm of wire at Q: $$F_Q =i_Ql B_{net}$$
$$F=\cfrac { 4\pi \times { 10 }^{ -7 }\times 20\times 10\times 10\times { 10 }^{ -2 } }{ 2\pi \times 10\times { 10 }^{ -2 } } $$
$$=\cfrac { 4\pi \times { 10 }^{ -7 }\times 10\times { 10 }^{ -2 } }{ 2\pi \times { 10 }^{ -2 } } \left[ 100-30 \right] $$
$$=20\times { 10 }^{ -7 }\times 70=1400\times { 10 }^{ -7 }$$
$$=1.4\times { 10 }^{ -4 }N$$ toward right.
A positively charged disk is rotated clockwise as shown in the figure. The direction of the magnetic field at point $$A$$ in the plane of the disk is
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$$\bigotimes $$ into the page
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$$\rightarrow $$ towards right
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$$\leftarrow $$
towards left
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$$\bigodot $$ out of the page
Explanation
Here, when we look at the rotating disc, we can treat is as a combination of infinite circular current carrying loops i.e. infinite infinitesimal dipoles.
As per the concept, the direction of the dipole will be such that the magnetic moment will go into the page (from South to North).
The point we are concerned about is in the equatorial plane of these dipoles and the the direction of magnetic field at this point will be From North to South ( as given in the concept ) i.e. out of the page.
Figure shows two long wires carrying equal currents $${I}_{1}$$ and $${I}_{2}$$ flowing in opposite directions. Which of the arrows labeled $$A,B,C$$ and $$D$$ correctly represents the direction of the magnetic field due to the wires at a point located at an equal distance $$d$$ from each wire?
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$$A$$
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$$B$$
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$$C$$
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$$D$$
Explanation
Magnitude of magnetic field at the given distance d from the wires is same because:
$$B_1=B_2=\dfrac{\mu_o I_1}{2 \pi d}$$
Net magnetic field, $$B=B_h+B_v$$
where horizontal component, $$B_h=B_1 \cos 30 - B_2 \cos 30 = 0$$
vertical component, $$B_v = B_1 \sin 30 + B_2 \sin 30 = B_1$$ upwards
Two very long, straight wires carrying currents as shown in figure. Find all locations where the net magnetic field is zero.
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$$y=\sqrt {2} x$$
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$$y=x$$
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$$y=-x$$
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$$y=-(x/2)$$
Explanation
Along the dashed line, $$\vec { { B }_{ 1 } } $$ and $$\vec { { B }_{ 2 }
} $$ are in opposite directions. If the line has slope $$-1.00$$,
then $${r}_{1}={r}_{2}$$ and $${B}_{1}={B}_{2}$$. So, $${B}_{net}=0$$
Hence, slope of the line dashed line is -1.
$$\Rightarrow y =-x$$ as the line passes through the origin making intercept zero.
A small current element of length $$dl$$ and carrying current is placed at $$(1, 1, 0)$$ and is carrying current in '$$+z$$' direction. If magnetic field at origin be $$\overrightarrow{B}_1$$ and at point $$(2, 2, 0)$$ be $$\vec B_2$$ then
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$$\overrightarrow{B} = \overrightarrow{B}$$
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$$|\overrightarrow{B}_1 | = | 2\overrightarrow{B}_2 |$$
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$$\overrightarrow{B}_1 = - \overrightarrow{B}_2$$
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$$\overrightarrow{B}_1 = - 2\overrightarrow{B}_2$$
Explanation
Now it is given that a small current carrying element is placed at $$(1,1,0)$$ and carrying current in $$+z$$ direction.
Then by using right hand thumb rule we can find the magnetic field lines which would be circular in this case having center at $$(1,1,0)$$
Now since origin and $$(2,2,0)$$ are two points which will lie on a circular centered at $$(1,1,0)$$ therefore, magnitude of magnetic field will be same at both the points.
And we find that origin and $$(2,2,0)$$ are diametrical opposite points therefore, direction of magnetic field at origin will be just opposite of the direction of magnetic field at $$(2,2,0)$$
Two thin long wires carry currents $${I}_{1}$$ and $${I}_{2}$$ along x-and y-axes respectively as shown in figure. Consider the points only in x-y plane.
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Magnetic field is zero at least at one point in each quadrant
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Magnetic field can be zero somewhere in the first quadrant
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Magnetic field can be zero somewhere in the second quadrant
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Magnetic field is non-zero in second quadrant
Explanation
In first and third quadrants, direction of magnetic field is opposite due to the two wires.
So, magnetic field can be zero only in first and third quadrants not in second and fourth.
Two infinitely long linear conductors are arranged perpendicular to each other and are mutually perpendicular planes as shown in figure. If $${I}_{1}=2A$$ along y-axis, $${I}_{2}=3A$$ along -ve z-axis and $$AP=AB=1cm$$, the value of magnetic field strength $$\vec { B } $$ at $$P$$ is
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$$\left( 3\times { 10 }^{ -5 }T \right) \hat { j } +\left( -4\times { 10 }^{ -5 }T \right) \hat { k } $$
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$$\left( 3\times { 10 }^{ -5 }T \right) \hat { j } +\left( 4\times { 10 }^{ -5 }T \right) \hat { k } $$
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$$\left( 4\times { 10 }^{ -5 }T \right) \hat { j } +\left( 3\times { 10 }^{ -5 }T \right) \hat { k } $$
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$$\left( -3\times { 10 }^{ -5 }T \right) \hat { j } +\left( 4\times { 10 }^{ -5 }T \right) \hat { k } $$
Explanation
$$\vec { { B }_{ 1 } } =\cfrac { { \mu }_{ 0 } }{ 2\pi (AP) } \hat { k }
=\cfrac { 4\pi \times { 10 }^{ -7 }\times 2 }{ 2\pi \times 1\times { 10
}^{ -2 } } \hat { k } =\left( 4\times { 10 }^{ -5 }T \right) \hat { k }
$$
Magnetic field strength at $$P$$ due the $${I}_{2}$$
$$\vec { {
B }_{ 2 } } =\cfrac { { \mu }_{ 0 } }{ 2\pi (BP) } \hat { j } =\cfrac {
4\pi \times { 10 }^{ -7 }\times 2 }{ 2\pi \times 2\times { 10 }^{ -2 } }
\hat { j } =\left( 3\times { 10 }^{ -5 }T \right) \hat { j } $$
Hence, $$\vec { B } =\left( 3\times { 10 }^{ -5 }T \right) \hat { j } +\left( 4\times { 10 }^{ -5 }T \right) \hat { k } $$
A coil carrying a heavy current and having large number of turns is mounted in a N-S vertical plane. A current flows in the clockwise direction. A small magnetic needle at its centre will have its north pole in
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east-north direction
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west-north direction
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east-south direction
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west-south direction
Explanation
When the plane of coil is in N-S vertical plane and current is passed in clockwise direction, the magnetic field at the centre of coil due to current is directed East to West. Earth's magnetic field is south to North.
Since the magnetic needle will be in the direction of resultant magnetic field, which will be in the west-north direction.
A wire of length $$L$$ metre carrying a current $$I$$ ampere is bent in the form of a circle. It's magnitude of magnetic moment will be
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$$\dfrac{IL}{ 4\pi}$$
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$$\dfrac{I^2 L^2}{4 \pi}$$
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$$\dfrac{IL^2 }{ 4 \pi}$$
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$$\dfrac{IL^2}{8 \pi}$$
Explanation
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Practice Class 12 Medical Physics Quiz Questions and Answers
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