A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure. A uniform magnetic field B exists in the perpendicular direction. Find the emf induced (a) in the loop abc, (b) in the segment bc, (c) in the segment ac and (d) in the segment ab.

Open in App

Solution

(a) The emf induced in loop abc is zero, as there is no change in the magnetic flux through it.
(b) The emf induced is given by
$e = (\vec{v} \times \vec{B}) . \vec{l}$
Emf induced in segment bc, e = Bvl (With positive polarity at point C)

(c) There is no emf induced in segment bc, as the velocity is parallel to its length.

(d) The emf induced in segment ab is calculated by the following formula:
e = B.v. (Effective length of ab)
The effective length of ab is along the direction perpendicular to its velocity.
Emf induced, e = B.v.(bc)

Suggest Corrections

Similar questions

Q. A metallic square loop is moving in its own plane with velocity

v

in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced

Q. An angle

a o b

made of a conducting wire moves towards right along its bisector through a uniform magnetic field

B

as shown in the figure. Find the net emf induced between two free ends if the magnetic field is perpendicular to the plane of the angle. Velocity of the angle is

v

Q. A conducting square loop of side

l

and resistance R moves in its plane with a uniform velocity

v

perpendicular to one of its sides. A uniform and constant magnitude field B exists along the perpendicular to the plane of the loop as shown in figure. The current induced in the loop is

Q. A wire-loop confined in a plane is rotated in its own plane with some angular velocity. A uniform magnetic field exists in the region. Find the emf induced in the loop.

Metallic square loop ABCD is moving in its own plane with velocity V in a uniform magnetic field perpendicular to its plane as shown in the fig. An electric field is induced

Join BYJU'S Learning Program