Question 199: Electromagnetic Induction

Question

A long rectangular conducting loop of width ' $l$ ', mass ' $m$ ' and resistance ' $R$ ' is placed partly in a perpendicular magnetic field ' B '. It is pushed downwards with velocity ' $V$ ' so that it may continue to fall freely. The velocity ' $V$ ' is

Accepted Answer

Step 1: Find the induced emf
When the loop moves in the magnetic field, an emf (electric force) is created. The formula is: $$ E = l v B $$ where $l$ is the width of the loop, $v$ is the velocity, and $B$ is the magnetic field.
Step 2: Find the induced current
According to Ohm's Law, $E = iR$, where $i$ is the current and $R$ is the resistance. So, $$ i = \frac{l v B}{R} $$
Step 3: Equate forces for free fall
If the loop is falling freely, the downward force (gravity) must equal the upward magnetic force. So, $$ mg = F_m $$ where $m$ is mass and $g$ is gravity.
The magnetic force ($F_m$) is given by $i l B$. So, $$ mg = i l B $$
Step 4: Substitute the current value
Substitute $i$ from step 2 into the force equation: $$ mg = \left(\frac{l v B}{R}\right) l B $$ $$ mg = \frac{l^2 v B^2}{R} $$
Step 5: Solve for velocity $v$
Rearrange for $v$: $$ v = \frac{mgR}{B^2 l^2} $$

Answer

$\frac{\mathrm{mgR}^2}{\mathrm{~B} l}$

Answer

$\frac{\mathrm{B}^2 l^2 \mathrm{R}}{\mathrm{mg}}$

Answer

$\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}$

Answer

$\frac{\mathrm{mgl}}{\mathrm{B}^2 \mathrm{R}^2}$

Electromagnetic Induction

A long rectangular conducting loop of width ' l ', mass ' m ' and resistance ' R ' is placed partly in a perpendicular magnetic field ' B '. It is pushed downwards with velocity ' V ' so that it may continue to fall freely. The velocity ' V ' is

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