Answer:
The magnetic field strength is 2.11 T.
Explanation:
It is given that,
Length of the wire, L = 37 cm = 0.37 m
Current, I = 3.2 A
Force, F = 2.5 N
Magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90,sin\ \theta=1[/tex]
[tex]F=qvB[/tex]
We know that,
Velocity, [tex]v=\dfrac{L}{t}[/tex]
[tex]F=\dfrac{qLB}{t}[/tex]
Since, [tex]\dfrac{q}{t}=I[/tex]
[tex]F=ILB[/tex]
[tex]B=\dfrac{F}{IL}[/tex]
[tex]B=\dfrac{2.5\ N}{3.2\ A\times 0.37\ m}[/tex]
B = 2.11 T
So, the magnetic field strength is 2.11 T. Hence, this is the required solution.
