The radius of the circular conducting loop shown in figure is R- Magnetic field is decreasing at a constant rate - Resistance per unit length of the loop is - Then current in wire AB is AB is one of the diametersxxxA- from A to BB- from B to A2 R - from A to BD- zero

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Motional EMF in an Open Conductor

The radius of...

Question

The radius of the circular conducting loop shown in figure is $R$ . Magnetic field is decreasing at a constant rate $α$ . Resistance per unit length of the loop is $ρ$ . Then current in wire $A B$ is ( $A B$ is one of the diameters)

\frac{R α}{2 ρ}

from

A

B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R α}{2 ρ}

from

B

A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 R α}{ρ}

from

A

B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $zero$
According to Lenz's law emf of the same magnitude in the clockwise direction are induced in the two loops into which the figure is divided. So, current is induced in the clockwise direction in the boundary but no current in wire $A B$ .

Suggest Corrections

Similar questions

Q. The radius of the circular conducting loop shown in Fig. is

R

. Magnetic field is decreasing at a constant rate

α

. Resistance per unit length of the loop is

ρ

.
Then, the current in wire

A B

is (

A B

is one of the diameters)

Q. Figure shows a uniform circular loop of radius

a

having resistance per unit length

ρ

placed in a unifrom magnetic field

B

perpendicular to plane of figure. A uniform rod of length

2 a

& resistance

R

moves with a velocity

v

as shown.
Find the current in the rod when it has moved a distance

\frac{a}{2}

from the centre of circular loop.

Given B = \sqrt{3}, a = 2, v = 3, ρ = \frac{9}{4 π}, R = \frac{2}{\sqrt{3}}

all in SI units.

Q. A circular conducting loop of radius

r_{0}

and having resistance per unit length

λ

is placed in a magnetic field

B

which is constant in space and time as shown in the figure. The ends of the loop are crossed and pulled in opposite directions with a velocity

v

such that the loop always remains circular and the radius of the loop goes on decreasing, then :

Q. The magnetic flux density

B

is changing its magnitude at a constant rate

d B / d t

. A given mass

m

of copper having resistivity

ρ

and density

d

, drawn into a wire of radius

a

and formed into a circular loop of radius

r

is placed perpendicular to the field

B

. The induced current in the loop

i

is :

Q. A conducting circular loop of radius

a

is connected to two long, straight wires. The straight wires carry a current

i

as shown in figure. Find the magnetic field

B

at the centre of the loop.

Join BYJU'S Learning Program

The radius of the circular conducting loop shown in figure is R. Magnetic field is decreasing at a constant rate α. Resistance per unit length of the loop is ρ. Then current in wire AB is (AB is one of the diameters)

The correct option is D zeroAccording to Lenz's law emf of the same magnitude in the clockwise direction are induced in the two loops into which the figure is divided. So, current is induced in the clockwise direction in the boundary but no current in wire AB.

The radius of the circular conducting loop shown in figure is $R$ . Magnetic field is decreasing at a constant rate $α$ . Resistance per unit length of the loop is $ρ$ . Then current in wire $A B$ is ( $A B$ is one of the diameters)

The correct option is D $zero$
According to Lenz's law emf of the same magnitude in the clockwise direction are induced in the two loops into which the figure is divided. So, current is induced in the clockwise direction in the boundary but no current in wire $A B$ .