Magnetic Effect of Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics

Each and every pair of loop elements located symmetrically w.r.t. central line experience zero net force, so total magnetic force experienced by the loop is zero.

As we know that, radius of a charged particle in a magnetic field $$B$$ is given by
$$r = \frac{{mv}}{{qB}}\,......\left( {\text{i}} \right)$$
where, $$r =$$  charge on the particle
$$v =$$  speed of the particle
$$\therefore $$ The time taken to complete the circle,
$$\eqalign{ & T = \frac{{2\pi r}}{v} \Rightarrow \frac{T}{{2\pi }} = \frac{m}{{qB}}\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr & \therefore \omega = \frac{{2\pi }}{T} = \frac{{qB}}{m} \cr & \because q = e\,\,{\text{and}}\,\,\frac{e}{m} = 1.76 \times {10^{11}}C/kg \cr & B = 3.57 \times {10^{ - 2}}T \cr & \Rightarrow \frac{{2\pi }}{T} = \frac{{eB}}{m}f = \frac{1}{{2\pi }}\frac{e}{m}B\,\,\left( {\because \frac{1}{T} = f} \right) \cr & = \frac{1}{{2\pi }} \times 1.76 \times {10^{11}} \times 3.57 \times {10^{ - 2}} \cr & = 1.0 \times {10^9}Hz = 1\,GHz \cr} $$

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).
Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by $$\frac{1}{2}m{v^2}$$  and $${v^2}$$ is the square of the magnitude of velocity which does not change.

Equating magnetic force to centripetal force,
$$\frac{{m{v^2}}}{r} = qvB\sin {90^ \circ }$$
Time to complete one revolution,
$$T = \frac{{2\pi r}}{v} = \frac{{2\pi m}}{{qB}}$$

Electric force on the particle, $$F = Eq,$$  and displacement $$s = \frac{1}{2}a{t^2} = \frac{1}{2}\left( {\frac{{Eq}}{m}} \right){t^2}.$$
Now, $$W = \Delta K,$$
$$\eqalign{ & {\text{or}}\,\,Fs = \frac{1}{2}m\left( {v_f^2 - v_i^2} \right)\,\,{\text{or}}\,\,Eq \times \frac{1}{2}\left( {\frac{{Eq}}{m}} \right){t^2} \cr & = \frac{1}{2}m\left[ {{{\left( {2{v_0}} \right)}^2} - v_0^2} \right] \cr & \therefore t = \frac{{\sqrt 3 m{v_0}}}{{qE}}. \cr} $$

Total internal resistance $$ = \left( {50 + 2950} \right)\,\Omega = 3000\,\Omega $$
Emf of the cell, $$\varepsilon = 3\,V$$
$$\therefore {\text{Current}} = \frac{\varepsilon }{R} = \frac{3}{{3000}} = 1 \times {10^{ - 3}}A = 1.0\,mA$$
$$\therefore $$ Current for full scale deflection of $$30$$ divisions is $$1.0\,mA.$$
$$\therefore $$ Current for a deflection of 20 divisions,
$$I = \left( {\frac{{20}}{{30}} \times 1} \right)mA\,\,{\text{or}}\,\,I = \frac{2}{3}mA$$
Let the resistance be $$x\,\Omega .$$ Then
$$x = \frac{\varepsilon }{I} = \frac{{3V}}{{\left( {\frac{2}{3} \times {{10}^{ - 3}}A} \right)}} = \frac{{3 \times 3 \times {{10}^3}}}{2}\Omega = 4500\,\Omega $$
But the resistance of the galvanometer is $$50\,\Omega .$$
$$\therefore $$ Resistance to be added $$ = \left( {4500 - 50} \right)\Omega $$
$$ = 4450\,\Omega $$

KEY CONCEPT : When a charged particle enters perpendicular to a magnetic field, then it moves in a circular path of radius.
$$r = \frac{p}{{qB}}$$
where $$q$$ = Charge of the particle
$$p$$ = Momentum of the particle
$$B$$ = Magnetic field
Here $$p, q$$  and $$B$$ are constant for electron and proton, therefore the radius will be same.

The ratio $$\frac{M}{L}$$ is always $$\frac{q}{{2m}}$$

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by $$\tau = NBi\,A\sin \theta .$$    Torque is maximum when $$\theta = {90^ \circ },$$  i.e., the plane of the coil is parallel to the field.
$${\tau _{\max }} = NBiA$$

Forces $${F_1}$$ and $${F_2}$$ acting on the coil are equal in magnitude and opposite in direction. As the forces $${F_1}$$ and $${F_2}$$ have the same line of action, their resultant effect on the coil is zero.
The two forces $${F_3}$$ and $${F_4}$$ are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is $$F,$$ the net force on the remaining three arms of the loop is $$-F.$$

The net force on the loop
$$\eqalign{ & {F_{{\text{net}}}} = {F_{AB}} + {F_{BC}} + {F_{CD}} + {F_{AD}} \cr & \Rightarrow {F_{{\text{net}}}} = \frac{{{\mu _0}iI}}{\pi } - \frac{{{\mu _0}iI}}{{3\pi }} = \frac{{2{\mu _0}iI}}{{3\pi }} \cr} $$