23.(b) A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, the wire is turned from N-S to northeast-northwest direction,

Answers (1)
S Sayak

Magnetic field strength =1.5 T.

Current flowing through the wire=7.0 A

The angle between the direction of the current and magnetic field=45o

The radius of the cylindrical region=10.0 cm

The length of wire inside the magnetic field, \\l=\frac{2r}{sin\theta }\\ 

Force on a wire in a magnetic field is calculated by relation,

\\F=BIlsin\theta \\ \\F=1.5\times 7\times \frac{2\times 0.1}{sin45^{o}}\times sin45^{o}

F=2.1 N

This force due to the magnetic field inside the cylindrical region acts on the wire in the vertically downward direction.

This force will be independent of the angle between the wire and the magnetic field as we can see in the above case.

Note: There is one case in which the force will be zero and that will happen when the wire is kept along the axis of the cylindrical region.

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