Name: RL Circuit
Uploaded: 2019-07-19
Duration: 10 min
Description: What happens when a resistor and an inductor are connected? How does it behave? We will learn the mathematical formula for the physics behind this circuit.

Question

Find the EMF induced in metal loop when it moves in non-uniform magnetic.

Accepted Answer

Define the direction of $v$ as horizontal direction and in the plane perpendicular to $v$ as upward direction.

Induced emf in a rod can be calculated by the following expression:
$E = (v \times B) . L$ . $L$ can be defined by joining the end points of the shape.

Direction of $v \times B$ can be defined by right-hand thumb rule. By right-hand rule, the direction of $v \times B$ is found in upward direction.

For rod $P Q$ and $R S$ , $L$ is along horizontal direction, so $v \times B$ and $L$ are perpendicular to each other and dot product of these two quantities will be $0$ .
So, for $P Q$ and $R S$ rods, $E = 0$ .

For rods $R P$ and $S Q$ , $L$ is along vertical direction, so $v \times B$ and $L$ are parallel to each other and dot product of these two quantities will be maximum, i.e. $E_{R P} = B_{R P} v L = B_{R P} v a$ and $E_{S Q} = B_{S Q} v L = B_{S Q} v a$ . Where, $B_{R P}$ and $B_{S Q}$ are magnetic fields at $R P$ and $S Q$ rod respectively. As the direction of emfs will be in upward direction in both the rods, the net emfs of the rods will be subtraction of both the emfs.

So, $E_{n e t} = E_{P R} - E_{S Q}$
$E_{n e t} = B_{R P} v a - B_{S Q} v a$
$E_{n e t} = \frac{μ_{0} I}{2 πx} v a - \frac{μ_{0} I}{2 π (x + a)} v a$
$E_{n e t} = \frac{µ_{0} I v a^{2}}{2 π x (x + a)}$

Find the EMF induced in metal loop when it moves in non-uniform magnetic.

Important Questions on Electromagnetic Induction

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