Moving Charges and Magnetism

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New answer posted•a year 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 :

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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

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

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New answer posted•a year 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

17 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Dynamic Resistance at 2v

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

Dynamic Resistance at 4 v

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

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

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New answer posted•a year 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.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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New answer posted•a year 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

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$B_{1} = \frac{μ_{0} i}{2 R}$

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

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

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New answer posted•a year 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).

37 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$B = 0.5 T$

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

= $i l B s i n 60$

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

= 0.216 N

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New answer posted•a year 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.

14 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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