A uniform wire of length 5m is carrying a steady current. The electric field inside it is 0.2 $v m^{- 1}$ . The potential difference across the ends of the wire is

1 v o l t

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

0.5 v o l t

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

0.1 v o l t

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

5 v o l t

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

Open in App

Solution

The correct option is A $1 v o l t$
We know,

$d V = E . d l$

$\int d V = \oint E . d l = E . \oint d l$

$V = E l = 5 \times 0.2 = 1 V$

Option $A$ is the correct answer

Suggest Corrections

Similar questions

A current 0. 2 A flows in a wire of resistance 15 $Ω$ . Find the potential difference across the ends of the wire.

When a potential difference of 2v is applied across the end of a wire of 5 m length a current 1 amp flows through it calculate

(a) the resistance across the ends of the wire if it is doubled on itself

When a potential difference of 2 V is applied across the ends of a wire of $5 m$ length, a current of $1 A$ flows through it. Calculate the resistance across the ends of the wire if its length is doubled.

A wire has a length of 2.0 m and a resistance of 5.0 $Ω$ Find the electric field existing inside the wire if it carries a current of 10 A.

Q. Consider a wire carrying a steady current placed in a uniform magnetic field B perpendicular to its length Consider the charges inside the wire It is known that magnetic forces do no work This implies that

Join BYJU'S Learning Program

A uniform wire of length 5m is carrying a steady current. The electric field inside it is 0.2 vm−1. The potential difference across the ends of the wire is

The correct option is A 1voltWe know,dV=E.dl∫dV=∮E.dl=E.∮dlV=El=5×0.2=1VOption A is the correct answer

A uniform wire of length 5m is carrying a steady current. The electric field inside it is 0.2 $v m^{- 1}$ . The potential difference across the ends of the wire is

The correct option is A $1 v o l t$
We know,

$d V = E . d l$

$\int d V = \oint E . d l = E . \oint d l$

$V = E l = 5 \times 0.2 = 1 V$

Option $A$ is the correct answer