Question

A wire of length Z has a linear mass density α, carrying current 'i', is suspended in the air where there is a uniform magnetic field \(\vec B \). The magnitude of the magnetic field is 
1. \(\frac{\alpha g}{i}\)
2. \(\frac{\alpha i}{g}\)
3. \(\frac{\alpha i^2}{g}\)
4. \(\frac{\alpha g^2}{i}\)

Answer 1

Correct Answer - Option 1 : \(\frac{\alpha g}{i}\)

CONCEPT:

• The flow of current in a conductor is due to the motion of free electrons in a definite direction.
  • When such a conductor is placed in a magnetic field, each electron moving in the uniform magnetic field experiences a force in a specific direction.
  • The direction of this force is given by fleming's left-hand rule.
• Let us take a conductor of uniform cross-section A and length l. Let the number density of mobile charge carriers (electron, in this case) be n. So total mobile no of mobile charge carriers will be nlA.
• In presence of an external magnetic field, force in these charge carriers is given by 

\(⇒ \overrightarrow{F}= nlAq \overrightarrow{v_{d}}× \overrightarrow{B}\)
Where q is the charge and B is the external magnetic field, |nqvd|A is the current I flowing through the conductor. 

Magnetic force in a current-carrying conductor is given by
\(⇒ \overrightarrow{F}= I \overrightarrow{l}× \overrightarrow{B}\)
Where \(\vec{F}\) is the magnetic force, \(\vec{B}\) is the magnetic field, \(\vec{l}\)is the length of the conductor, and I is the current flowing through the conductor, and θ is the angle between the length of the conductor (in direction of current flow) and magnetic field.

EXPLANATION:

Given: θ, = 90°, length of the conductor = Z, mass per unit length = α

• Magnetic force in a current-carrying conductor is given by

\(\Rightarrow F = Il B\sin θ \)

As the wire remain suspended in air, therefore the weight of fire = force exerted by the magnetic field

\(\Rightarrow mg= iZB\sin90^0=iZB\) or \(α Zg=iZB\)

\(\Rightarrow B=\frac{ α g}{i}\)

Hence option 1) is correct