A uniform magnetic field exists in region which forms an equilateral triangle of side a . The magnetic field is perpendicular to the plane of the triangle. A charge q enters into this magnetic field perpendicularly with speed v along perpendicular bisector of one side and comes out along the perpendicular bisector of other side. The magnetic induction in the triangle is

Question

A uniform magnetic field exists in region which forms an equilateral triangle of side a . The magnetic field is perpendicular to the plane of the triangle. A charge q enters into this magnetic field perpendicularly with speed v along perpendicular bisector of one side and comes out along the perpendicular bisector of other side. The magnetic induction in the triangle is (A) (m v/q a) (B) (2 m v/q a) (C) (m v/2 q a) (D) (m v/4 q a)

Tardigrade Admin · Accepted Answer

The charged particle moves in a circle of radius (a/2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ec404ec05cacc8015441de30002eb6cb-.png" /> ∴ qvB=(m v2/(a)2) or B=(2 m v/q a)

$\frac{m v}{q a}$

$\frac{2 m v}{q a}$

$\frac{m v}{2 q a}$

$\frac{m v}{4 q a}$

The charged particle moves in a circle of radius $\frac{a}{2}$

$∴ q v B = \frac{m v ^{2}}{\frac{a}{2}}$ or $B = \frac{2 m v}{q a}$

qamv​

qa2mv​

2qamv​

4qamv​

The charged particle moves in a circle of radius 2a​ ∴qvB=2a​mv2​ or B=qa2mv​

$\frac{m v}{q a}$

$\frac{2 m v}{q a}$

$\frac{m v}{2 q a}$

$\frac{m v}{4 q a}$

The charged particle moves in a circle of radius $\frac{a}{2}$

$∴ q v B = \frac{m v ^{2}}{\frac{a}{2}}$ or $B = \frac{2 m v}{q a}$