Question

A horizontal length of current-carrying wire is suspended from two identical flexible leads that are under tension due to the weight of the wire. The wire is oriented at right angles to a uniform magnetic field that has a magnitude of 4T and is directed out of the screen of the figure. The length L of the wire is 5 cm and its mass mis 10 g. What current will produce a magnetic force on the wire such that it triples the tension in the flexible leads, and which way is the current flowing through the wire? Current-carrying wire suspended from identical flexible leads Ceiling BO magnetic field (out of the screen) 1, current-carrying wire L O 0.33 A, current flows right to left O 0.33 A, current flows left to right 0.98 A, current flows right to left O 0.98 A, current flows left to right 0.67 A, current flows left to right

Answer 1

Given: Initially let the total tension be T balancing the weight of the wire.

m = 10 g = 0.01 kg

B = 4 T

L = 5 cm = 0.05 m

As T balances weight of wire:

T = mg----(i)

when current i flows through the wire, the tension is tripled due to the additional magnetic force on the wire, and this magnetic force must be in the direction of weight of the wire.

therefore, 3T = mg + F

=> F = 3T - mg

=> F = 3mg - mg = 2mg-----(ii)

Therefore, the magnitude of magnetic force is 2mg.

Also, the magnetic force on a current carrying wire is given by

F = iLB-----(iii)

Therefore, from (ii) and (iii),

iLB = 2mg

=> i = 2mg/LB

$\frac{2*0.01*9.81}{0.05*4}=0.981A = 0.98A$.[answer]

The vector form of magnetic force on current carrying wire is $\vec{F}=i(\vec{L}\times \vec{B})$ .

The direction of F is downward, the direction of magnetic field B is out of the screen.

therefore, by the application of right hand thumb rule, we get the direction of current from left to right. [answer]

The correct option is 4th option.