Question

A 20 cm long wire carrying a current of 3.0 ampere is kept at an angle of 30° with the magnetic field of 5.0 N /A- m . How much force will act on wire ? If wire is kept perpendicular to the magnetic field, then how much force will act?

Answer 1

Answer:

F= BIL sin(theta)

F = 5 X 0.5 X 3 X 0.2

F = 1.5 Newton

when sin(theta) = 1

F = 5 X 0.2 X 3

3 Newton

Answer 2

Answer:

A 20 cm long wire carrying a current of 3A is kept at an angle of 30° with a magnetic field of 5.0 N /A- m. The force on the wire is $1.5N$, If the wire is kept perpendicular to the magnetic field, then the force will be $3N$

Explanation:

When a wire carries current $I$ is kept in a magnetic field of intensity $B$ it will experience a force $F=IlBsin$θ

from the question, the current through the wire is $I=3A$

length of the wire is$l=20cm=0.20m$

the magnetic field strength $B=5N/Am$

the wire is kept by making an angle of $30$⁰ with the magnetic field

substitute the given values to get force

$F=3*0.20*5 sin30=1.5N$

if the wire is kept perpendicular to the field then θ=$90$⁰

now the force is $F=3*0.20*5=3N$