A block of mass m and charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform and out of the plane of the paper as shown. Find the time from the start when block leaves contact with the surface.

Question

A block of mass m and charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform and out of the plane of the paper as shown. Find the time from the start when block leaves contact with the surface. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-rxzreohqkzvljyvx.jpg" /> (A) (m cos θ /q B) (B) (m cos e c θ /q B) (C) (m cot θ /q B) (D) (m tan θ /q B)

Tardigrade Admin · Accepted Answer

Block will lose contact with surface when force due to magnetic field will become equal to

m g cos θ

q v B = m g cos θ

v = \frac{m g c o s θ}{qB} = u + a t

( along inclined plane )

v = \frac{m g c o s θ}{qB} = (g sin θ) t

t = \frac{m cot θ}{qB}

Q. A block of mass $m$ and charge $q$ is released on a long smooth inclined plane. Magnetic field $B$ is constant, uniform and out of the plane of the paper as shown. Find the time from the start when block leaves contact with the surface.

$\frac{m c o s θ}{qB}$

$\frac{m c o s ec θ}{qB}$

$\frac{m c o t θ}{qB}$

$\frac{m t a n θ}{qB}$

Block will lose contact with surface when force due to magnetic field will become equal to $m g cos θ$
$q v B = m g cos θ$
$v = \frac{m g c o s θ}{qB} = u + a t$ ( along inclined plane )
$v = \frac{m g c o s θ}{qB} = (g sin θ) t$
$t = \frac{m cot θ}{qB}$

Q. A block of mass m and charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform and out of the plane of the paper as shown. Find the time from the start when block leaves contact with the surface.

qBmcosθ​

qBmcosecθ​

qBmcotθ​

qBmtanθ​