Question
Write the expression for the force on a charge moving in a magnetic field.Use this expression to define the SI unit of magnetic field.
✓
Answer
Force on a charge (q) moving in a magnetic field B with velocity $\vec{\text{v}}$ making an angle $\theta$ (with the direction of magnetic field $(\vec{\text{B}})$ is given by,$\text{F}_\text{m}=\text{qvB}\sin\theta$
when $\theta=90^\circ\Rightarrow\sin\theta=1,$so
$\text{F}_\text{m}=\text{qvB}$
$\text{or}\ \text{B}=\frac{\text{F}_\text{m}}{\text{q}^\text {v}}$
$\text{If}\ \text{v}=1\ \text{m/s},\text{B}=\frac{\text{F}_\text{m}}{\text{q}}\ \text{newton/ coulomb}$
SI unit of magnetic field is telsa.
Thus, 1 tesla is the magnetic field in which a charged particle moving with velocity 1m/ s perpendicular to velocity experiences a force of 1 newton/ coulomb.
when $\theta=90^\circ\Rightarrow\sin\theta=1,$so
$\text{F}_\text{m}=\text{qvB}$
$\text{or}\ \text{B}=\frac{\text{F}_\text{m}}{\text{q}^\text {v}}$
$\text{If}\ \text{v}=1\ \text{m/s},\text{B}=\frac{\text{F}_\text{m}}{\text{q}}\ \text{newton/ coulomb}$
SI unit of magnetic field is telsa.
Thus, 1 tesla is the magnetic field in which a charged particle moving with velocity 1m/ s perpendicular to velocity experiences a force of 1 newton/ coulomb.
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