Question

The force acting on a wire of length 500m. Carrying 35A current placed perpendicular magnetic field of 2 Tesla.​

Answer 1

Answer :-

$\implies \boxed{ \sf{ F = 35000 \: N \: {m}^{2}}} \:$

To Find :-

→ Force acting on wire .

Explanation :-

Given that

• Length of wire (L) = 500m

• current (i) = 35 Ampere

• magnetic field (B) = 2 Tesla

• Magnetic Force (F) = ?

We know that ,

$\implies \boxed{\sf{ F = iLB\sin\theta }\: }$

here Force is perpendicular hence angle be 90° .

$\implies \sf{ F =35 \times 500 \times 2 \times \sin 90 \degree} \\ \\ \implies \boxed{ \sf{ F = 35000 \: N \:{m}^{2}}}$

Direction taken by right hand screw rule .

→ Thumb show direction of current and fingers show direction of magnetic field .

Answer 2

$\tt{\huge{\orange {Hello!!! }}}$

QUESTION :

Find The force acting on a wire of length 500m, Carrying 35A current placed perpendicular magnetic field of 2 Tesla.

ANSWER :

The required force is 35000 N.

SOLUTION :

We have the following information given in the question :

The length of the wire is 500 m.

The current flowing in it is 35 Amperes.

The magnetic field is 2 Tesla.

Required Formulae :

Force = A × L × B × Sin { phi }

Where ,

A refers to the current flowing in the wire.

B refers to the magnetic field .

L refers to the length of the wire .

Using the Right Hand Thumb Rule, We can state that :

Phi = 90°

=> Sin { Phi } = 1

Now substituting the values into the above Equation,

F = 35 × 500 × 2 × 1 N = 35000 N.......(A)