Question 208: Electromagnetic Induction

Question

A coil of ' $n$ ' turns and area ' $A$ ' is suddenly removed from a magnetic field, a charge ' $q$ ' flows through the coil. If resistance of the coil is ' $R$ ' then the magnetic flux density is (in $\mathrm{Wb} / \mathrm{m}^2$ )

Accepted Answer

Step 1: FaradayÃ¢ÂÂs law relation
The induced emf (total across coil):
$ \mathcal{E} = -n \frac{d\Phi}{dt} $
where $\Phi = BA$ is the flux through one turn (if field is uniform and perpendicular).
Step 2: Relation between emf and current
$ \mathcal{E} = I R $
Step 3: Charge flow
Total charge that flows over the discharge interval:
$ q = \int I \, dt = \frac{1}{R} \int \mathcal{E}\, dt $
Step 4: Evaluate the integral
$ \int \mathcal{E}\, dt = -n \int d\Phi = n\,\Delta\Phi $
Here, change in flux per turn is initial $BA$ to zero, so $\Delta\Phi = BA$.
Thus:
$ q = \frac{1}{R}(n BA) $
Step 5: Solve for $B$
$ B = \frac{q R}{n A} $
Final Answer:
$ \boxed{\frac{qR}{nA}} $
This corresponds to
Option B
.

Answer

$\frac{q^2 R}{2 n A}$

Answer

$\frac{q R}{n A}$

Answer

$\frac{q R^2}{n A}$

Answer

$\frac{q R}{2 n A}$

Electromagnetic Induction

A coil of ' n ' turns and area ' A ' is suddenly removed from a magnetic field, a charge ' q ' flows through the coil. If resistance of the coil is ' R ' then the magnetic flux density is (in \mathrm{Wb} / \mathrm{m}^2 )

Choose the correct answer: