The magnetic field \( B \) is initially given by the equation:
Step 1: \( B = \frac{\mu_0 i}{2R} \times 4 \), which is the original magnetic field expression.
Step 2: For the new radius \( R' = 4R \), the new field \( B' \) is given by:
Step 3: Substitute \( R' \) into the magnetic field formula: \( B' = \frac{\mu_0 i}{2R'} = \frac{\mu_0 i}{8R} \).
Step 4: Now, calculate the ratio \( \frac{B'}{B} = \frac{1}{16} \), indicating that the new magnetic field is \( \frac{1}{16} \) of the original field.
Step 5: With this ratio, we conclude that the new magnetic field \( B' \) is \( 2T \).
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.}
Let $ [.] $ denote the greatest integer function. If $$ \int_1^e \frac{1}{x e^x} dx = \alpha - \log 2, \quad \text{then} \quad \alpha^2 \text{ is equal to:} $$
If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:
Three distinct numbers are selected randomly from the set $ \{1, 2, 3, ..., 40\} $. If the probability that the selected numbers are in an increasing G.P. is $ \frac{m}{n} $, where $ \gcd(m, n) = 1 $, then $ m + n $ is equal to:
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where