First slide

Force on a current carrying wire placed in a magnetic field

Question

A horizontal rod of mass 10 g and length 10 cm is placed on a smooth plane inclined at an angle of $60^{0}$ with the horizontal with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 ampere, the value of B for which the rod remains stationary on the inclined plane is

Moderate

A

1.73 tesla
B

(1/1.73) tesla
C

1 tesla
D

0.5 tesla

Solution

force on the conductor due magnetic field IlB along horizontal direction and its component along the plane is ilB cos $θ$ which when balances mgsin $θ$ it will be in equilibrium
$m g \sin θ = i l B \cos θ B = \frac{m g}{i l} \tan θ$

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