A uniform magnetic field of $1.5\; T$ exists in a cylindrical region of radius $10.0\; cm$, 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,
$(a)$ the wire intersects the axis,
$(b)$ the wire is turned from $N-S$ to northeast-northwest direction,
$(c)$ the wire in the $N-S$ direction is lowered from the axis by a distance of $6.0 \;cm?$
A uniform magnetic field of $1.5\; T$ exists in a cylindrical region of radius $10.0\; cm$, 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,
$(a)$ the wire intersects the axis,
$(b)$ the wire is turned from $N-S$ to northeast-northwest direction,
$(c)$ the wire in the $N-S$ direction is lowered from the axis by a distance of $6.0 \;cm?$
Magnetic field strength, $B =1.5\, T$
Radius of the cylindrical region, $r=10 \,cm =0.1 \,m$
Current in the wire passing through the cylindrical region, $I=7\, A$
$(a)$ If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, $I=2 r=0.2 \,m$
Angle between magnetic field and current, $\theta=90^{\circ}$
Magnetic force acting on the wire is given by the relation, $F=B I l \sin \theta$
$=1.5 \times 7 \times 0.2 \times \sin 90^{\circ}$
$=2.1\, N$
Hence, a force of $2.1 \,N$ acts on the wire in a vertically downward direction.
$(b)$ New length of the wire after turning it to the northeast-northwest direction can be given
$l_{1}=\frac{l}{\sin \theta}$
Angle between magnetic field and current, $\theta=45^{\circ}$ Force on the wire,
$F=B I_{1} \sin \theta$
$=B I I$
$=1.5 \times 7 \times 0.2$
$=2.1 \,N$
Hence, a force of $2.1\,N$ acts vertically downward on the wire. This is independent of angle $\theta$ because $l \sin \theta$ is fixed.
$(c)$ The wire is lowered from the axis by distance, $d=6.0 \,cm$
Let $l_{2}$ be the new length of the wire.
$\therefore\left(\frac{l_{2}}{2}\right)^{2}=4(d+r)$
$=4(10+6)=4(16)$
$\therefore l_{2}=8 \times 2=16 \,cm =0.16 \,m$
Magnetic force exerted on the wire, $F_{2}=B I I_{2}$
$=1.5 \times 7 \times 0.16$
$=1.68\, N$
Hence, a force of $1.68\, N$ acts in a vertically downward direction on the wire.
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