
Formulas for motional emf \((E = Blv)\) and the force on a current carrying conductor in a magnetic field \((F = BIl)\). Make sure your units are con sistent
The force on a conductor in a magnetic field is given by:
\( F = I \ell B \)
Where:
The current \( I \) can be expressed as:
\( I = \frac{e}{R} \)
Substitute \( I \) into the force equation:
\[ F = Bv \ell B \cdot \frac{\ell}{R} \]
Simplify:
\[ F = \frac{B^2 \ell^2 v}{R} \]
Given:
Substitute these values into the equation:
\[ F = \frac{(15)^2 \cdot (1)^2 \cdot 4}{5} \]
Simplify:
\[ F = \frac{225 \cdot 4}{5} = 180 \, \text{N} \]
The magnetic force is \( F = 18 \, \text{N}. \)
A bar magnet has total length \( 2l = 20 \) units and the field point \( P \) is at a distance \( d = 10 \) units from the centre of the magnet. If the relative uncertainty of length measurement is 1\%, then the uncertainty of the magnetic field at point P is:
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 
An infinite wire has a circular bend of radius \( a \), and carrying a current \( I \) as shown in the figure. The magnitude of the magnetic field at the origin \( O \) of the arc is given by:


For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: