Question

A straight horizontal metal wire carries 28A current. What magnitude of magnetic field is needed to balances
weight? Its linear mass density is 56 g/m
(1) 8.2 mT
(2) 11.2 mT
(3) 19.6 mT
(4) 25.4 mT​

Answer 1

Answer:

19.6 mT

Explanation:

F = BiL

From above formula

B = F / (iL)

= mg / (iL)

= (m/L) × g/i

= 56 g/m × (9.8 m/s²)/(28 A)

= 0.056 kg/m × (9.8 m/s²)/(28 A)

= 0.0196 T

= 19.6 mT

Answer 2

The magnitude of magnetic field is needed to balances weight is 19.6 mT (Option 3 is the correct answer)

Given,

A straight horizontal metal wire carries 28A current

linear mass density is 56 g/m

To Find,

Find the magnitude of magnetic field is needed to balances weight

Solution,

• A magnetic field is produced around a wire when an electric current flows through it. Additionally, the wire is encircled by concentric rings created by this magnetic field. Additionally, the current's direction determines the magnetic field's direction.
• Additionally, we may figure it out by applying the "right-hand rule," which involves pointing the thumb of your right hand towards the direction of the current. Additionally, the magnetic field aligns up with the direction of your curled fingers.
• Additionally, the size of the field is influenced by the current flowing through it as well as its distance from the wire delivering the charge.

From the formula of magnitude of magnetic field we know that,

$B= \frac{F}{iL}$

where, The force on a section of wire of length L carrying a current I through a magnetic field B is

so,

$B= \frac{m*g}{i*L}$

$B= (\frac{m}{L} )*(\frac{g}{i} )$

$B= \frac{56*9.8}{28}$

$B= 19.6 mT$

Hence, the magnitude of magnetic field is needed to balances weight is 19.6 mT

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