Question

Help me out in this question:)
If F=qvBsin theta, where F is force, q is charge, v is velocity, B is magnetic field and sin theta is the angle between v and B, then find the dimension of B.

Answer 1

Answer:

F=qvB Sin@

here @ is angle say theta...

now B= F/(qvSin@).............(i)

Here F is force having dimensions

${m}^{1} {l}^{1} {t}^{ - 2}$

q is charge having dimensions

${a}^{1} {t}^{ 1}$

v is velocity having dimensions

${l}^{1} {t}^{ - 1}$

Sin@ is trigonometric ratio it is dimension less....

So net dimension of B (mag. field) is given by

$\frac{( {m}^{1} {l}^{1} {t}^{ - 2} )}{( {a}^{1} {t}^{1} ) \times ( {l}^{1} {t}^{ - 1}) } \\ = = = \: \: \: {m}^{1} {a}^{1} {t}^{ - 2}$

where a is ampeare [A]

m is mass [M]

l is length [L]

t is time [T]

And SI unit of B (mag. field) is Tesla (T)

$\large{Hope\:it\:helps\:you}$