Question

A 20-cm long conducting wire of 5-g carrying 5-A current is placed in a uniform magnetic field. The strength and direction of the magnetic field needed to levitate the wire are

49.05 mT out of the page.
15 mT into the page.
49.05 mT into the page.
25 mT into the page.
25 mT out of the page.
references

Answer 1

\[F=BIL\]
\[F=mg\]
equate both of them and solve
Make sure you write all quantities in SI unit

B-->Magnetic Field

Answer 2

L->Length of wire (in m)
I-->Current in wire (through Ampere)

Answer 3

and g=9.8 wright

Answer 4

Yes

Answer 5

to levitate the wire the field should exert a force on it and it is givenF=iBl if angle between direction of current and magnetic field is 90 degrees

Answer 6

is that correct . 49.05 into the page using right hand rule

Answer 7

No You should be using left hand rule

Answer 8

i guess both A and D should be correct becauese in these cases the magnetic force is always gonna suuport mg and there will be net downard acc.@shivam_bhalla what do u say abt that?

Answer 9

yes yes sorry i want to write left

Answer 10

Siddharth18 . what do you mean about that!!!! do you hav explain

Answer 11

sorry options A and E(the last one)

Answer 12

According to left hand rule , the force should balance mg . Therefore, the force should act in upward direction and for that the field should be only into the paper. If the field is acting out of paper, then it will not balance mg

Answer 13

hey guys i got it .the wire won't move when it is in equilibrium.so we just calculate B when the wire is in equilibrium and then all options excluding the one calculated at eqi will be correct