The current density in a cylindrical wire of radius R = 2 mm is uniform over the cross-section of the wire and is given by J = 2 × 10 4 A / m 2 . Find the current through the outer portion of the wire between the radial distance R 2 and R .

Given: The current density J in a cylinderical wire of radius R = 2 mm= 2 × 10 - 3 m is 2 × 10 4 A / m 2 . To find: The current through the outer portion of the wire between the radial distance R 2 and R . The cross-section area A enclosed between the radius R 2 and R is given by: A = π [ R 2 - ( R 2 ) 2 ] = 3 4 πR 2 , on substituting the value, we get, A = 3 4 πR 2 = 3 4 × 3 . 14 × ( 2 × 10 - 3 ) 2 m 2 Also, the current enclosed by the given cross section area will be: I = J × A = 2 × 10 4 × 3 4 × 3 . 14 × ( 2 × 10 - 3 ) 2 = 0 . 19 A

The current density in a cylindrical wire of radius R = 2 mm is uniform over the cross-section of the wire and is given by J = 2 × 10 4 A / m 2 . Find the current through the outer portion of the wire between the radial distance R 2 and R .

Given: The current density J in a cylinderical wire of radius R = 2 mm= 2 × 10 - 3 m is 2 × 10 4 A / m 2 . To find: The current through the outer portion of the wire between the radial distance R 2 and R . The cross-section area A enclosed between the radius R 2 and R is given by: A = π [ R 2 - ( R 2 ) 2 ] = 3 4 πR 2 , on substituting the value, we get, A = 3 4 πR 2 = 3 4 × 3 . 14 × ( 2 × 10 - 3 ) 2 m 2 Also, the current enclosed by the given cross section area will be: I = J × A = 2 × 10 4 × 3 4 × 3 . 14 × ( 2 × 10 - 3 ) 2 = 0 . 19 A

Find the resistance of 0 . 5 Kg copper wire of diameter 0 . 008 m . Given that the density of copper = 8 . 9 × 10 3 kg / m 3 . Use its value of resistivity from Table 5.1 Given: Table 5.1

Given: A 0 . 5 Kg Copper wire of diameter 0 . 008 m . Density of copper = 8 . 9 × 10 3 kg / m 3 Resistivity from Table 5.1 of the book is 1 . 724 × 10 - 8 ohm - m To find the resistance of the wire. The cross-section area A = πD 2 4 = π ( 8 × 10 - 3 ) 2 4 m 2 Volume of the wire = M a s s d e n s i t y = 0 . 5 8 . 9 × 10 3 = 0 . 056 × 10 - 3 m 3 The length of the wire l = v o l u m e a r e a = 0 . 056 × 10 - 3 π ( 8 × 10 - 3 ) 2 4 = 1 . 1 m So the resistance of the wire R = ρ l A = 1 . 724 × 10 - 8 1 . 1 π ( 8 × 10 - 3 ) 2 4 = 3 . 83 × 10 - 4 Ω

An electric lamp has a filament diameter of 0 . 525 mm . For 2 . 84 ampere of current in the lamp, 1 . 37 × 10 22 electrons / m 3 are found to flow through the wire, Calculate the average speed of the electrons in it.

Given: An electric lamp has a filament diameter of 0 . 525 mm . The current I in the lamp 2 . 84 ampere The current density n = 1 . 37 × 10 22 electrons / m 3 To calculate: The average speed V d of the electrons in it. Area of cross-section A of the filament = πD 2 4 = π ( 0 . 525 × 10 - 3 ) 2 4 m 3 Applying formulae, I = n e A V d Putting the values, we get, 2 . 84 = 1 . 37 × 10 22 × 1 . 6 × 10 - 19 × π ( 0 . 525 × 10 - 3 ) 2 4 V d o r V d = 2 . 84 1 . 37 × 10 22 × 1 . 6 × 10 - 19 × π ( 0 . 525 × 10 - 3 ) 2 4 = 18 . 8 × 10 3 m / s

HARD

12th Odisha Board

IMPORTANT

Earn 100

The current density in a cylindrical wire of radius $R = 2 mm$ is uniform over the cross-section of the wire and is given by $J = 2 \times 10^{4} A / m^{2}$ . Find the current through the outer portion of the wire between the radial distance $\frac{R}{2}$ and $R$ .

Important Questions on Electric Current

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 13

(a) The current density in a cylindrical wire of radius $R = 2 mm$ is uniform over the cross-section of the wire and is given by $J = 2 \times 10^{4} A / m^{2}$ . Find the current through the outer portion of the wire between the radial distance $\frac{R}{2}$ and $R$ .

(b) If $J = f (r)$ and is given by $J = f (r)$ where $J = b r^{2}$ where $b = 3 \times 10^{10} \frac{A}{m}$ and $r$ is any radial distance in metres, find the current through the given portion as in(a)

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 14

The copper winding of a motor have a resistance of $50 Ω$ at $20^{°} C$ , when the motor is idle. After running for several hours, the resistance rises to $58 Ω$ . Calculate the temperature of the windings at the end. $α_{C u} = 3.93 \times 10^{- 3} K^{- 1}$ .

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 15

A wire with a resistance of $6 Ω$ is drawn out so that its new length is thrice its original length. Find the resistance of the longer wire assuming that the resistivity and density of the material do not change during the process.

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 16

Find the resistance of $0.5 Kg$ copper wire of diameter $0.008 m$ . Given that the density of copper $= 8.9 \times 10^{3} kg / m^{3}$ . Use its value of resistivity from Table 5.1

Given: Table 5.1

Question Image

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 17

A battery connected to a rheostat furnishes certain current. When a lamp of resistance $150 Ω$ is inserted in series, the current is reduced to one third of its former value. Find the resistance of the rheostat.

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 18

An electric lamp has a filament diameter of $0.525 mm$ . For $2.84$ ampere of current in the lamp, $1.37 \times 10^{22}$ electrons $/ m^{3}$ are found to flow through the wire, Calculate the average speed of the electrons in it.

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 19

A cylindrical metallic wire is stretched to increase its length by $10 %$ . Calculate the percentage increase in resistance.

HARD

12th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 12>Chapter 5 - Electric Current>MODEL QUESTIONS>Q 20

What is the change in resistance of an Eureka wire when its radius is halved and the length is reduced to one fourth of its original value.

The current density in a cylindrical wire of radius R=2 mm is uniform over the cross-section of the wire and is given by J=2×104 A/m2. Find the current through the outer portion of the wire between the radial distance R2 and R.

Important Questions on Electric Current

Learn and Prepare for any exam you want !

The current density in a cylindrical wire of radius $R = 2 mm$ is uniform over the cross-section of the wire and is given by $J = 2 \times 10^{4} A / m^{2}$ . Find the current through the outer portion of the wire between the radial distance $\frac{R}{2}$ and $R$ .