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Moving Charges and Magnetism

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2 months ago

Moving Charges and Magnetism Class 12th

A closely wounded circular coil of radius 5cm produces a magnetic field of 37.68 * 10^-4 T at its centre. The current through the coil is_ A. [Given, number of turns in the coil is 100 and π = 3.14]

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Vishal Baghel

Contributor-Level 10

$B_{c e n t r e} = N \frac{μ_{0} i}{2 r}$

$\Rightarrow \frac{100 * 4 π * 1 0^{- 7} * i}{2 * 5 * 1 0^{- 2}} = 37.68 * 1 0^{- 4}$

$∴ i = \frac{37.68}{4 * 3.14} = 3 A$

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2 months ago

Moving Charges and Magnetism Class 12th

The magnitude field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance $\frac{r}{2}$ from the centre is :

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Vishal Baghel

Contributor-Level 10

$B_{C} = \frac{μ_{0} l}{2 r} = B$ $_{} \frac{_{}}{}$ -(1)

$\begin{array}{l} B_{P} = \frac{μ_{0} l r^{2}}{2 {(r^{2} + \frac{r^{2}}{4})}^{3 / 2}} \\ B_{P} = \frac{μ_{0} l r^{2}}{2 * r^{3} {(\frac{5}{4})}^{3 / 2}} \end{array}$

$B_{P} = \frac{μ_{0} l}{2 r {(\frac{5}{4})}^{3 / 2}}$

$\Rightarrow B_{P} = \frac{B 4^{3 / 2}}{5^{3 / 2}}$

$B_{P} = B \frac{2^{3}}{{(\sqrt[]{5})}^{3}}$

$B_{P} = {(\frac{2}{\sqrt[]{5}})}^{3} B$

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2 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

The I-V characteristics of p-n junction diode in forward bias is shown in the figure. The ration of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is

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Payal Gupta

Contributor-Level 10

Dynamic Resistance at 2v

= $\frac{Δ v}{Δ I} = \frac{2.1 - 2}{(10 - 5) * 1 0^{- 3}} = \frac{0.1}{5} * 1 0^{3}$

Dynamic Resistance at 4 v

= $\frac{Δ v}{Δ I} = \frac{4.2 - 4}{(250 - 200)} = \frac{0.2}{50 * 1 0^{- 3}} = \frac{0.2}{50} * 1 0^{3}$

$\frac{R_{2 v}}{R_{4 v}} = \frac{0.1 * 1 0^{3} * 50}{5 * 0.2 * 1 0^{3}} = 5 : 1$

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3 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

An infinitely long hollow conducting cylinder with radius R carries a uniform current along its surface. Choose the correct representation of magnetic field (B) as a function of radial distance (r) from the axis of cylinder.

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Payal Gupta

Contributor-Level 10

Using Ampere's law for long hollow cylinder carrying current on its surface

B_in = 0

$B_{o u t} = \frac{μ_{0} i}{2 π r}$

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3 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

The magnetic field at the centre of current carrying circular loop is B₁. The magnetic field at a distance of $\sqrt[]{3}$ times radius of the given circular loop from the center on its axis is B₂. The value of B₁/B₂ will be

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Vishal Baghel

Contributor-Level 10

$B_{1} = \frac{μ_{0} i}{2 R}$

$B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + {(\sqrt[]{3} R)}^{2})}^{3 / 2}} = \frac{μ_{0} i R^{2}}{2 * 8 R^{3}} = \frac{μ_{0} i}{8 * 2 R} = \frac{B_{1}}{8}$

$\Rightarrow \frac{B_{1}}{B_{2}} = 8 : 1$

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3 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A triangular shaped wire carrying 10 A current is a uniform magnetic field of 0.5T, as shown in figure.

The magnetic force on segment CD is (Given BC = CD = BD = 5cm).

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Vishal Baghel

Contributor-Level 10

$B = 0.5 T$

$F_{C D} = i \vec{l} * \vec{B}$

= $i l B s i n 60$

$= 10 * \frac{5}{100} * 0.5 * \frac{\sqrt[]{3}}{2} = \frac{\sqrt[]{3}}{8} = \frac{1.732}{8}$

= 0.216 N

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3 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A singly ionized magnesium atom (A = 24) ion is accelerated to kinetic energy 5 ke V, and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be ________cm.

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Payal Gupta

Contributor-Level 10

$r = \frac{m v}{q B} = \frac{\sqrt[]{2 k m}}{q B}$

= $\frac{\sqrt[]{2 * 5 * 1 0^{3} * 1.6 * 1 0^{- 19} * 24 * 1.66 * 1 0^{- 27}}}{1.6 * 1 0^{- 19} * \frac{1}{2}}$

= 9.975 * 102 cm

= 9.975 cm

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3 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If r_d and r_p are the radii of their circular paths respectively, then the ratio $\frac{r_{d}}{r_{p}}$ will be $\sqrt[]{x} : 1$ where x is……….

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Payal Gupta

Contributor-Level 10

r_d = (Radius of deuteron)

$= \frac{m_{d} v_{d}}{q_{d} B}$

$= \frac{\sqrt[]{2 m_{d} k . E}}{q_{d} B}$

$= \frac{\sqrt[]{2 m_{P}}}{\sqrt[]{m_{P}}} * \frac{q_{P}}{q_{d}}$ $[q_{P} = q_{d} = 1.6 * 1 0_{C}^{- 19}]$

$= \sqrt[]{2} : 1$

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New answer posted

3 months ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two long parallel conductors S₁ and S₂ are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure).

A charge particle of 3p coulomb is passing through the point P with velocity $\vec{v} = (2 \hat{i} + 3 \hat{j})$ m/s; where $\hat{i} a n d \hat{j}$ represents unit vector along x & y axis respectively. The force acting on the charge particle is 4p * 10^-5 $(- x \hat{i} + 2 \hat{j}) N .$ The value of x is

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Payal Gupta

Contributor-Level 10

${\vec{B}}_{N e t} = \frac{μ_{0}}{4 π} * 2 [\frac{i_{1}}{d_{1}} - \frac{i_{2}}{d_{2}}] (- \hat{k})$

$= 1 0^{- 7} * 2 [\frac{4}{4} * 100 - \frac{2}{6} * 100] (- \hat{k})$

$= 2 * 1 0^{- 5} [1 - \frac{1}{3}] - \hat{k}$

${\vec{B}}_{N e t} = \frac{4}{3} * 1 0^{- 5} (- \hat{k})$

q = 3πc

$\vec{v} = 2 \hat{i} + 3 \hat{j}$

$\vec{F} = q (\vec{v} * \vec{B})$

= $3 π [(2 \hat{i} * 3 \hat{j}) * \frac{4}{3} * 1 0^{- 5} (- \hat{k})]$

$\vec{F} = 4 π * 1 0^{- 5} (- 3 \hat{i} + 2 \hat{j}) N$

x = 3

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4 months ago

Moving Charges and Magnetism Physics Mechanical Properties of Solids

The bulk modulus of a liquid is 3 × 1010 Nm-2. The pressure required to reduce the volume of liquid by 2% is

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