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9 months ago

Physics Class 12th

An EM wave propagating in x-direction has wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:

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alok kumar singh

Contributor-Level 10

Given, $λ = 8 m m, v = 8 m m, v = 3 * 1 0^{8}$

$\vec{E} = ε_{0} s i n [k x - ω t] \hat{j}$

$k = \frac{2 π}{λ} = \frac{2 π}{8 * 1 0^{- 3}} = \frac{π}{4} * 1 0^{3}$

$\vec{E} = 60 s i n [k (x - \frac{ω}{k} t)] \hat{j}$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - v t]] \hat{j} [? \frac{ω}{k} = v]$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{j} v / m$

Now $\frac{E_{0}}{B_{0}} = C$

B0 = $\frac{60}{3 * 1 0^{8}} = \frac{20}{1 0^{8}} = 2 * 1 0^{- 7}$ $\vec{B} = 2 * 1 0^{- 7} [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{k} T$

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9 months ago

Physics Physics Moving Charges and Magnetism

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 * 10^-6N, then the value of x is approximately:

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alok kumar singh

Contributor-Level 10

$\frac{F}{l} = \frac{μ_{0}}{2 π} \frac{i_{1} i_{2}}{d}$

$2 * 1 0^{- 6} = \frac{4 π * 1 0^{- 7}}{2 π} * \frac{x * x}{d}$

x² = 2x = $\sqrt[]{2}$ = 1.4x = 1.4

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9 months ago

Physics Class 12th

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:

(Assume the coil to be short circuited.)

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alok kumar singh

Contributor-Level 10

$P = \frac{E^{2}}{R} = \frac{{(N A \frac{d B}{d t})}^{2} * 4 A C}{ρ l}$

$p' = {(\frac{N A}{2} \frac{d B}{d t})}^{2} * 4 A C$

$p' = 2 P$

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9 months ago

Physics Physics Moving Charges and Magnetism

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 * 10^-5. What is fractional increase in the magnetic field inside solenoid with respect to air medium inside the solenoid?

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alok kumar singh

Contributor-Level 10

$χ (S u s c e p t i b i l i t y) = 1.2 * 1 0^{- 5}$

fractional change = $= \frac{Δ B}{B} = \frac{B i n s i d e - B o u t}{B o u t}$

$= \frac{μ_{m} H - μ_{0} H}{μ_{0} H} = \frac{(μ_{0} μ_{r} - μ_{0}) H}{μ_{0} H}$ $[? μ_{m} = b_{0} μ_{0}]$

$= μ_{r} - 1$

= $= (1 + χ) - 1 = χ = 1.2 * 1 0^{- 5}$

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9 months ago

Physics Class 12th

Resistance of the wire is measured as at 10°C and 30°C respectively. Temperature co-efficient of resistance of the material of the wire is:

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alok kumar singh

Contributor-Level 10

R_T = R₀ $(1 + α Δ T)$

at t = 10°C, R_T = 2 $Ω$

2 = R₀ (1 + [10 – T0])- (1)

at T = 30°C, R_T= 3 $Ω$

3 = $R_{0} [1 + α (30 - T_{0})]-$ (2)

from (1)

$2 = R_{0} (1 + α 10)-$ (3)

from (2) 3 = R₀ (1 + 30) - (4)

$(4) \div (3)$

$\frac{3}{2} = \frac{1 + 30 α}{1 + 10 α}$

3 + 30 = 2T 60α = $\frac{1}{30} = 0.033$

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9 months ago

Physics Class 12th

Two point charges A and B of magnitude + 8 * 10^-6 and -8 * 10^-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 * 10⁴ NC^-1. The distance 'd' between the point charges A and B is:

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

Enet = E₁ + E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$\Rightarrow 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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Physics Class 12th

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k ? q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

Enet = E₁+ E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$? 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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9 months ago

Physics Class 11th

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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9 months ago

Physics Class 11th

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

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alok kumar singh

Contributor-Level 10

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

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Physics Physics Moving Charges and Magnetism

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