Magnetic Effects of Current and Magnetism — Easy Practice (Set 1)

Quick Recap — Magnetic Field of Currents

• Oersted: a current produces a magnetic field; its direction follows the right-hand thumb rule.
• Standard fields: straight wire $B = \dfrac{\mu_0 I}{2\pi r}$; centre of a loop $B = \dfrac{\mu_0 I}{2R}$; solenoid $B = \mu_0 nI$ (with $\mu_0 = 4\pi\times10^{-7}$ T m/A).
• Ampere's law relates $\oint\vec{B}\cdot d\vec{l}$ to the enclosed current.
• Lorentz force: $\vec{F} = q\vec{v}\times\vec{B}$ ($F = qvB\sin\theta$); it does no work, so the speed stays constant.
• Charge in a field ($v\perp B$): a circular path of radius $r = \dfrac{mv}{qB}$ and period $T = \dfrac{2\pi m}{qB}$ (independent of speed).

Beyond-NCERT JEE Formulae

The definitions ($\vec F=q\vec v\times\vec B$, Ampere's law, and $B=\dfrac{\mu_0 I}{2\pi r}$) are taken as known from the practice sets. This sheet collects the higher-yield results and the extended finite-geometry formulae that JEE Main tests beyond the NCERT core, each with a note on when it applies.

1. Biot-Savart law and standard field sources

• Biot-Savart (vector form): $d\vec B=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}$. Reach for this whenever the geometry is an arc, a finite segment or a polygon, i.e. too unsymmetric for Ampere's law.
• Long straight wire: $B=\dfrac{\mu_0 I}{2\pi r}$, the lines circling the wire by the right-hand grip.
• Finite straight wire (perpendicular distance $r$; the ends subtend $\theta_1$ and $\theta_2$ at the foot of the perpendicular): $B=\dfrac{\mu_0 I}{4\pi r}(\sin\theta_1+\sin\theta_2)$. A semi-infinite wire ending level with the point gives $\dfrac{\mu_0 I}{4\pi r}$, and the infinite wire recovers $\dfrac{\mu_0 I}{2\pi r}$.
• Centre of a circular coil of $N$ turns: $B=\dfrac{\mu_0 N I}{2R}$, while a bare arc of angle $\phi$ contributes $\dfrac{\mu_0 I\phi}{4\pi R}$.
• On the axis of a loop, a distance $x$ from the centre: $B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$, which becomes $\dfrac{\mu_0 I}{2R}$ at the centre and dies off as $\dfrac{1}{x^3}$ far away.
• Long solenoid (interior): $B=\mu_0 nI$ with $n$ the turns per metre; it drops to exactly $\dfrac{\mu_0 nI}{2}$ at each open end.
• Toroid (inside the core): $B=\dfrac{\mu_0 N I}{2\pi r}$, and it is zero everywhere outside.

[JEE Tip] The finite-wire formula is the master key for polygon loops: every side of a regular loop shares the same perpendicular distance and the same pair of half-angles, so the centre field of a square of side $a$ works out to $\dfrac{2\sqrt2\,\mu_0 I}{\pi a}$. Learn the route, not the number.

2. Forces on currents

• On a wire: $\vec F=I\vec L\times\vec B$, of magnitude $BIL\sin\theta$. In a uniform field a bent or curved wire feels exactly the force of the straight wire joining its ends (the effective length $\vec L_{eff}$), so a closed loop feels zero net force.
• Between parallel wires: $\dfrac{F}{L}=\dfrac{\mu_0 I_1 I_2}{2\pi d}$; parallel currents attract and antiparallel currents repel, the opposite of the rule for like charges.
• Useful constants: $\dfrac{\mu_0}{2\pi}=2\times10^{-7}$ and $\dfrac{\mu_0}{4\pi}=10^{-7}$ in SI units.

[JEE Tip] Effective length is a uniform-field shortcut only. Near another current-carrying wire the field varies as $\dfrac{1}{r}$, so it is non-uniform and you must integrate rather than collapse the wire to its chord.

3. Charged particle in a magnetic field

• Circular motion ($\vec v\perp\vec B$): radius $r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}=\dfrac{\sqrt{2mqV}}{qB}$; period $T=\dfrac{2\pi m}{qB}$; cyclotron frequency $f=\dfrac{qB}{2\pi m}$. The last two are independent of speed and radius, which is exactly what makes the cyclotron work.
• Helical motion (velocity at angle $\theta$ to $\vec B$): only $v_\perp=v\sin\theta$ bends the path, so $r=\dfrac{mv\sin\theta}{qB}$, while $v_\parallel=v\cos\theta$ drives the particle forward one pitch $p=v_\parallel T=\dfrac{2\pi m v\cos\theta}{qB}$ per turn.
• Velocity selector (crossed $\vec E$ and $\vec B$): a charge crosses undeflected only at $v=\dfrac{E}{B}$, independent of $q$ and $m$, which is why it heads every mass spectrometer.
• Cyclotron exit energy: $K_{max}=\dfrac{q^2 B^2 R^2}{2m}$, fixed entirely by the dee radius $R$ and the field $B$.

[JEE Tip] The magnetic force does no work, so $v$ and $K$ never change, only the direction does. Equal-$K$ particles then compare by $r\propto\dfrac{\sqrt m}{q}$, but equal-$V$ particles by $r\propto\sqrt{\dfrac{m}{q}}$; confusing the two wrecks isotope and $e/m$ problems.

4. Magnetic dipole: moment, torque and energy

• Moment of a coil: $M=NIA$ (units A m$^2$), pointing along the normal by the right-hand rule.
• Torque in a field: $\vec\tau=\vec M\times\vec B$, of magnitude $MB\sin\theta$, largest when the coil's plane contains $\vec B$.
• Orientation energy: $U=-\vec M\cdot\vec B=-MB\cos\theta$, so the external work to reorient a dipole is $W=MB(\cos\theta_1-\cos\theta_2)$.
• Bar magnet as a solenoid / short dipole: a uniformly magnetised bar behaves as a solenoid of the same moment $M=NIA$; its far field is axial $B=\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ and equatorial $B=\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$ (axial twice equatorial, both $\propto\dfrac{1}{r^3}$), and its small oscillations in a field obey $T=2\pi\sqrt{\dfrac{I}{MB}}$ with $I$ the moment of inertia.
• From a moving charge (beyond NCERT): an orbiting charge has the gyromagnetic ratio $\dfrac{M}{L}=\dfrac{q}{2m}$, and a charge $q$ smeared round a ring of radius $R$ spun at $\omega$ carries $M=\dfrac{q\omega R^2}{2}$.

[JEE Tip] Torque is not energy: turn the coil with $MB\sin\theta$, but budget the work with $(\cos\theta_1-\cos\theta_2)$. And a loop in a uniform field feels a torque yet zero net force, so it rotates on the spot without drifting.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Field at the centre and on the axis of a coil

Q. A flat circular coil of $100$ turns and radius $5$ cm carries a steady current of $2$ A. Find the magnetic field at its centre and at a point on the axis $5$ cm from the centre, and compare the two.

Solution. At the centre all $N$ turns add, so $B_c=\dfrac{\mu_0 N I}{2R}$: $$B_c=\frac{(4\pi\times10^{-7})(100)(2)}{2(0.05)}=\frac{2.513\times10^{-4}}{0.1}.$$ So $B_c=2.51\times10^{-3}$ T. On the axis, $B_x=\dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$; putting $x=R$ makes $(R^2+x^2)^{3/2}=2\sqrt2\,R^3$, so $B_x=\dfrac{B_c}{2\sqrt2}$: $$B_x=\frac{2.51\times10^{-3}}{2\sqrt2}=\frac{2.51\times10^{-3}}{2.828}.$$ So $B_x=8.89\times10^{-4}$ T, only $\dfrac{1}{2\sqrt2}\approx0.354$ of the central value just one radius off-axis. Once you set $x=R$ the ratio needs no re-substitution.

Example 2 — Field at the centre of a square loop

Q. A wire carries a steady current of $5$ A once around a square loop of side $20$ cm. Using the finite-wire result, find the magnetic field at the centre of the square.

Solution. Each side is a finite straight wire at perpendicular distance $r=\dfrac{a}{2}=0.1$ m, and its two ends each subtend $45^\circ$ at the centre, so $\sin\theta_1+\sin\theta_2=2\sin45^\circ=\sqrt2$. One side therefore gives $$B_{side}=\frac{\mu_0 I}{4\pi (a/2)}\sqrt2=\frac{\sqrt2\,\mu_0 I}{2\pi a}.$$ All four sides drive the field the same way at the centre, so they simply add: $$B=4B_{side}=\frac{2\sqrt2\,\mu_0 I}{\pi a}=2\sqrt2\times\frac{(4\times10^{-7})(5)}{0.2}.$$ So $B=2\sqrt2\times10^{-5}=2.83\times10^{-5}$ T. The value drops straight out of the finite-wire law; the circular-loop formula does not apply to a polygon.

Example 3 — Force between two parallel wires

Q. Two long parallel wires $5$ cm apart carry steady currents of $10$ A and $15$ A in the same direction. Find the force per unit length between them and its nature, and the total force on a $2$ m length.

Solution. With $\dfrac{\mu_0}{2\pi}=2\times10^{-7}$, the force per metre is $\dfrac{F}{L}=\dfrac{\mu_0 I_1 I_2}{2\pi d}$: $$\frac{F}{L}=\frac{(2\times10^{-7})(10)(15)}{0.05}=\frac{3\times10^{-5}}{0.05}.$$ So $\dfrac{F}{L}=6\times10^{-4}$ N/m, and because the currents are parallel the wires attract. Over a $2$ m length the total force is $$F=\left(\frac{F}{L}\right)L=(6\times10^{-4})(2).$$ So $F=1.2\times10^{-3}$ N. Reversing one current keeps this magnitude but turns the pull into a push.

Example 4 — Radius, period and frequency of a proton

Q. A proton ($m=1.67\times10^{-27}$ kg, $q=1.6\times10^{-19}$ C) moves at $2\times10^{6}$ m/s at right angles to a uniform field of $0.5$ T. Find the radius, period and frequency of its circular path.

Solution. The magnetic force is centripetal, so $r=\dfrac{mv}{qB}$: $$r=\frac{(1.67\times10^{-27})(2\times10^{6})}{(1.6\times10^{-19})(0.5)}=\frac{3.34\times10^{-21}}{8\times10^{-20}}.$$ So $r=4.18\times10^{-2}$ m, about $4.2$ cm. The period is $T=\dfrac{2\pi m}{qB}$: $$T=\frac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.5)}=\frac{1.049\times10^{-26}}{8\times10^{-20}}.$$ So $T=1.31\times10^{-7}$ s, and the frequency is $f=\dfrac{1}{T}=\dfrac{qB}{2\pi m}=7.62\times10^{6}$ Hz, about $7.6$ MHz. Doubling the speed would double the radius but leave $T$ and $f$ unchanged.

Example 5 — Velocity selector, then bending

Q. In a velocity selector a $3\times10^{5}$ V/m electric field and a $0.2$ T magnetic field act at right angles to each other and to a proton beam. Find the selected speed. The emerging protons then enter a $0.5$ T field at right angles; find the radius of their path.

Solution. Undeflected passage needs the electric and magnetic forces to cancel, $qE=qvB_1$, so $v=\dfrac{E}{B_1}$, independent of charge and mass: $$v=\frac{3\times10^{5}}{0.2}.$$ So $v=1.5\times10^{6}$ m/s. In the second field the magnetic force is centripetal, so $r=\dfrac{mv}{qB_2}$: $$r=\frac{(1.67\times10^{-27})(1.5\times10^{6})}{(1.6\times10^{-19})(0.5)}=\frac{2.505\times10^{-21}}{8\times10^{-20}}.$$ So $r=3.13\times10^{-2}$ m, about $3.1$ cm. The selector sets the speed and the second field does the bending, so never reuse $B_1$ in the radius.

Example 6 — Torque, energy and work on a coil

Q. A coil of $50$ turns and area $8\times10^{-3}$ m$^2$ carries $3$ A in a uniform field of $0.5$ T. Find its magnetic moment, the maximum torque, the torque when the moment is $30^\circ$ from the field, and the work to turn it from alignment to $60^\circ$.

Solution. The moment is $M=NIA=(50)(3)(8\times10^{-3})$, so $M=1.2$ A m$^2$. The torque $\tau=MB\sin\theta$ is greatest when the coil's plane lies along $\vec B$ (so $\sin\theta=1$): $$\tau_{max}=MB=(1.2)(0.5).$$ So $\tau_{max}=0.6$ N m. At $\theta=30^\circ$ the torque falls to $\tau=MB\sin30^\circ=(0.6)(0.5)$, giving $\tau=0.3$ N m. Using $U=-MB\cos\theta$, the work to swing the coil from $\theta_1=0$ to $\theta_2=60^\circ$ is $$W=MB(\cos0^\circ-\cos60^\circ)=(0.6)(1-0.5).$$ So $W=0.3$ J. The turning factor $\sin\theta$ and the energy factor $(1-\cos\theta)$ are different, so do not swap them.

Example 7 — Radius and pitch of a helix

Q. An electron ($m=9.1\times10^{-31}$ kg, $q=1.6\times10^{-19}$ C) enters a uniform field of $1\times10^{-3}$ T at a speed of $3\times10^{6}$ m/s, its velocity making $30^\circ$ with the field. Find the radius and the pitch of the resulting helix.

Solution. Split the velocity into $v_\perp=v\sin30^\circ=1.5\times10^{6}$ m/s, which bends the path, and $v_\parallel=v\cos30^\circ=2.60\times10^{6}$ m/s, which carries it along the field. The radius uses $v_\perp$, so $r=\dfrac{mv_\perp}{qB}$: $$r=\frac{(9.1\times10^{-31})(1.5\times10^{6})}{(1.6\times10^{-19})(10^{-3})}=\frac{1.365\times10^{-24}}{1.6\times10^{-22}}.$$ So $r=8.53\times10^{-3}$ m, about $8.5$ mm. One full turn takes $T=\dfrac{2\pi m}{qB}=3.57\times10^{-8}$ s, and in that time $v_\parallel$ advances the pitch: $$p=v_\parallel T=(2.60\times10^{6})(3.57\times10^{-8}).$$ So $p=9.28\times10^{-2}$ m, about $9.3$ cm. Only $v_\perp$ fixes the radius, while $v_\parallel$ alone sets the pitch.