Question

A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?

Answer 1

Answer:

Angle the wire make with respect to the magnetic field is 30°.

Explanation:

It is given that,

Length of wire, L = 0.6 m

Current flowing in the wire, I = 2 A

Magnetic field strength, B = 0.3 T

It is placed in the magnetic field. It will experience a force of, F = 0.18 N. We need to find the angle the wire make with respect to the magnetic field. The force acting on the wire is given by :

$F=I(L\times B)$

$F=ILB\ sin\theta$

$\theta=sin^{-1}(\dfrac{F}{ILB})$

$\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\ m\times 0.3\ T})$

$\theta=30^{\circ}$

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.