Direction of current in the loop at \$t=\dfrac{\pi}{3 \omega}\$ :

\$B = B_o sin(wt)\$\$\cfrac{d B}{dt} = B_ow cos(wt) \$at \$t = \cfrac{\pi}{3 \omega}\$\$\cfrac{d B}{dt} = B_ow/2 \$The magnetic field is increasing.According to Lenz's law, the current is induced to resist the change in magnetic field.Hence According to right hand law current is in anticlockwise direction.

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Magnetism And Matter

Q: An electric motor is a device to convert electrical energy into mechanical energy. The motor shown below has a rectangular coil \$(15\ cm \times 10\ cm)\$ with 100 turns placed in a uniform magnetic fleld \$B=2.5\ T\$. When a current is passed through the coil, lt completes \$50\$ revolutions in one second. the power output of the motor is \$1.5\ kW\$. The current rating of the coil should be

Let \$I\$ be the current through the coil of 100 turns. Let the magnetic moment \$\vec{M}\$ make an angle \$\theta \$ at an instant of time. Then, torque on the coil \$|\vec{\tau }|=MB\sin\theta \$Work done by the torque in each half revolution \$=\dfrac{W}{2} \$, where \$W\$ is the work done in one full revolution.\$\therefore \dfrac{W}{2}=\int_{0}^{\pi }MBsin\theta d\theta =IANB\int_{0}^{\pi }\sin\theta d\theta \$\$=2IANB\$\$\therefore \$work output per revolution \$=W=4INAB\$Hence work output per second\$=Power=4INABf\$\$\therefore 1.5\times 10^{3}=4INABf\$\$I=\dfrac{1.5\times 10^{3}}{4\times 100\times 150\times 10^{-4}\times 2.5\times 50}\$\$\therefore I=\dfrac{15\times 10^{2}}{15\times 50}=2\ A\$

Q: A bar magnet of magnetic moment \$M\$ is divided into '\$n\$' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength \$2T\$ and held making an angle \$60^{0}\$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:

Answer is ANet magnetic moment after cutting\$= \dfrac{M}{n}\$We know that, \$ |U|= MB \cos \theta\$In equilibrium position, all the potential energy will be converted into kinetic energy i.e. \$ KE=|U| \$Given \$ M_{new}=\dfrac{M}{n}, B=2T, \theta=60^{\circ} \$Therefore, \$K.E= \dfrac{M}n\times 2\times \cos 60^{\circ}=\dfrac{M}{n} J \$

Q: The tangent drawn at any point on the magnetic lines of force gives direction of mangetic field at that point.

True Magnetic lines of forces are imaginary lines in electric field which indicates the direction of magnetic field. Tangent at any point on a line of force give the direction of magnetic field at that point.

Q: A planar coil of area 7 \$\mathrm{m}^{2}\$ carrying an anti-clockwise current 2 A is placed in an extemal magnetic field \$\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})\$, such that the normal to the plane is along the line\$(3\hat{i}-5\hat{j}+4\hat{k})\$. Select correct statements from the following . ( Consider Normal of the coil and Magnetic moment vectors to be in the same direction )

Area \$A=7m^{2},\$ Normal \$\hat{n}=\dfrac{1}{\sqrt{50}}(3\hat{i}-5\hat{j}+4\hat{k})\$\$\simeq \dfrac{1}{7}(3\hat{i}-5\hat{j}+4\hat{k})\$\$\therefore \$ area \$\vec{A}=(3\hat{i}-5\hat{j}+4\hat{k})m^{2}\$Magnetic field \$\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})T\$Current \$I=2A\$\$\therefore \$ magnetic moment \$\vec{M}=I\vec{A}=(6\hat{i}-10\hat{j}+8\hat{k})Am^{2}\$\$|\vec{M}|\simeq 14Am^{2}(2\times 7=14)\$Potential energy U\$=-\vec{M}.\vec{B}\$\$=-[1.2-2-2.4]=3.2J\$Let \$\theta \$ be the angle between \$\hat{n}\$ and \$\vec{B}\$(i.e.,)between \$\vec{M}\$ and \$\vec{B}.\$Now,U\$=-MBcos\theta .\$\$\therefore cos\theta =-\dfrac{U}{MB}=-\dfrac{3.2}{14\times 0.4}(\because \sqrt{0.17}\simeq 0.4)\$\$=-\dfrac{4}{7}=-0.57\$\$\theta =\pi -cos^{-1}(0.57)\$

Q: The figure shows an isosceles triangle wire frame with apex angle equal to \$\pi /2\$. Theframe starts entering into the region of uniform magnetic field \$B\$ with constant velocity\$v\$ at \$t= 0\$. The longest side of the frame is perpendicular to the direction of velocity.If i is the instantaneous current through the frame then choose the alternative showingthe correct variation of i with time

The rate of change of area in the field starts from zero as the tip enters the field and grows linearly with time until it reaches the maximum value as the last slice enters; after that the whole loop is in the field so the flux through the loop is constant and there is no induced emf. hence the correct option is (d)

Q: \$N = 300\$ turns of this wire are uniformly wound on a permanent magnet shaped as a cylinder whose length is equal to \$l = 15\ cm\$. When a current \$I = 3.0\ A\$ was passed through the wiring the field outside the magnet disappeared. Find the coercive force \$H_{c}\$ of the material from which the magnet was manufactured.

The coercive force \$H_c\$ is just the magnetic field within the cylinder. This is by circulation theorem, \$\displaystyle H_c=\frac{NI}{l}=6\:kA/m\$(from \$\displaystyle\oint{\vec{H}\cdot d\vec{r}}=I\$, total current, considering a rectangular contour.)

Q: In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields \$B_1\$ (solid line) and \$B_2\$ (dashed line). If \$B_2\$ is larger than \$B_1\$, which of the following graphs shows the correct variation of R and T in these fields?

Resistance usually varies directly with temperature.But when considering superconductors, the least temperature is Tc(0).In presence of B, Tc decreases. But that doesn't say anything about general R and T relation.Only info we have is Tc(\$B_2\$) B_1\\). (from observing the graph)

Q: A thin iron ring with mean diameter \$d = 50\ cm\$ supports a winding consisting of \$N = 800\$ turns carrying current \$I = 3.0\ A\$. The ring has a cross-cut of width \$b = 2.0\ mm\$. Neglecting the scattering of the magnetic flux at the gap edges, and using the plot shown in Fig., find the permeability of iron under these conditions.

From the theorem on circulation of vector \$\vec{H}\$,\$\displaystyle H\pi d+\frac{Bb}{\mu_0}=NI\$ or, \$\displaystyle B=\frac{\mu_0NI}{b}-\frac{\mu_0\pi d}{b}H=(1.51-0.987)H\$,Where \$B\$ is in Tesla and \$H\$ in \$kA/m\$. Besides, \$B\$ and \$H\$ are interrelated as in the figure shown below of the text. Thus we have to solve for \$B\$, \$H\$ graphically by simultaneously drawing the two curves (the hysterisis curve and the straight line, given above) and find the point of intersection. It is at\$H\approx0.26\:kA/m\$, \$B=1.25\:T\$Then \$\displaystyle\mu=\frac{B}{\mu_0H}\approx4000\$.

Physics
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JEE Mains JEE Advanced NEET AIIMS BITSAT NTSE Easy Qs Med Qs Hard Qs

An electric motor is a device to convert electrical energy into mechanical energy. The motor shown below has a rectangular coil $(15 c m \times 10 c m)$ with 100 turns placed in a uniform magnetic fleld $B = 2.5 T$ . When a current is passed through the coil, lt completes $50$ revolutions in one second. the power output of the motor is $1.5 k W$ . The current rating of the coil should be

1 A

2 A

3 A

4 A

Hard

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The magnetic lines of forces :

Interested each other

Do not intersect each other

Parallel to each other

Perpendicular to each other

Hard

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A bar magnet of magnetic moment $M$ is divided into ' $n$ ' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength $2 T$ and held making an angle $6 0^{0}$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:

\frac{M}{n}

M n

\frac{M}{n ^{2}}

M n^{2}

Hard

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The tangent drawn at any point on the magnetic lines of force gives direction of mangetic field at that point.

True

False

Hard

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A planar coil of area 7 $m^{2}$ carrying an anti-clockwise current 2 A is placed in an extemal magnetic field $B = (0.2 \hat{i} + 0.2 \hat{j} - 0.3 \hat{k})$ , such that the normal to the plane is along the line $(3 \hat{i} - 5 \hat{j} + 4 \hat{k})$ . Select correct statements from the following . ( Consider Normal of the coil and Magnetic moment vectors to be in the same direction )

The potential energy of the coil in the given orientation is

6.4 J

The angle between the normal (positive) to the coil and the external magnetic field is

cos^{- 1} (0.57)

The potential energy of the coil in the given orientatlon is

3.2 J

The magnitude of magnetic moment of the coil is about 14

A m^{2}

Hard

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The figure shows an isosceles triangle wire frame with apex angle equal to $π / 2$ . The
frame starts entering into the region of uniform magnetic field $B$ with constant velocity
$v$ at $t = 0$ . The longest side of the frame is perpendicular to the direction of velocity.
If i is the instantaneous current through the frame then choose the alternative showing
the correct variation of i with time

Hard

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$N = 300$ turns of this wire are uniformly wound on a permanent magnet shaped as a cylinder whose length is equal to $l = 15 c m$ . When a current $I = 3.0 A$ was passed through the wiring the field outside the magnet disappeared. Find the coercive force $H_{c}$ of the material from which the magnet was manufactured.

Hard

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In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields $B_{1}$ (solid line) and $B_{2}$ (dashed line). If $B_{2}$ is larger than $B_{1}$ , which of the following graphs shows the correct variation of R and T in these fields?

Hard

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A thin iron ring with mean diameter $d = 50 c m$ supports a winding consisting of $N = 800$ turns carrying current $I = 3.0 A$ . The ring has a cross-cut of width $b = 2.0 m m$ . Neglecting the scattering of the magnetic flux at the gap edges, and using the plot shown in Fig., find the permeability of iron under these conditions.

Hard

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Direction of current in the loop at $t = \frac{π}{3 ω}$ :

clockwise

anticlockwise

no current will flow at that time

cannot be determined

Hard

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Magnetism And Matter

Physics 1375 Views

The magnetic lines of forces :

The tangent drawn at any point on the magnetic lines of force gives direction of mangetic field at that point.

In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1​ (solid line) and B2​ (dashed line). If B2​ is larger than B1​, which of the following graphs shows the correct variation of R and T in these fields?

Direction of current in the loop at t=3ωπ​ :

Physics
1375 Views

Direction of current in the loop at $t = \frac{π}{3 ω}$ :