Question

A straight copper wire of length 1m weighing 50 grams is carrying a current of 24.5A.it is held horizontally, in magnetic field, the magnitude of the magnetic field required to suspend the wire
A.1.0T
B.0.02T
C.0.2T
D.2.0T

Answer 1

Explanation:

Here, m=500g=0.5kg,I=4A,l=2.5m

As F=IlBsinθ

As the rod is suspended in air so,weight of thr rod is balanced by the magnetic force acting on the current carrying rod

mg=IlBsin90

o

, (∵θ=90

o

andF=mg)

∴B=

Il

mg

=

4×2.5

0.5×10

=0.5T

Answer 2

length of the wire(l)= 1m

mass of the wire=50 g

current flowing through the wire(i)= 24.5A

acceleration due to gravity=g

the weight of the copper wire is balanced by the magnetic field acting on the current carrying copper wire

F=BIL sin 90

F=BIL

mg=BIL

B=mg/i ×L

B=0.05×9.8/24.5×1

B=0.02T

I hope this helps ( ╹▽╹ )