Question

Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uniform magnetic field. the segment makes an angle of 53° with the direction of the magnetic field. if the force on the segment is 0.20 n, what is the magnitude of the magnetic field?

Answer 1

The force exerted by a magnetic field on a wire carrying current is:
$F=ILB \sin \theta$
where I is the current, L the length of the wire, B the magnetic field intensity, and $\theta$ the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is $53 ^{\circ}$, so we can re-arrange the formula and substitute the numbers to find B:
$B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T$

Answer 2

Answer:

The force exerted by a magnetic field on a wire carrying current is:

where I is the current, L the length of the wire, B the magnetic field intensity, and  the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is , so we can re-arrange the formula and substitute the numbers to find B: