Ratio of B at centre of circular coil radius \(r\) to B at distance \(r\) on axis is \(\sqrt{x}:1\). Find \(x\):
A4
B8
C16
D2
✅ Correct: B
\(B_\text{centre} = \dfrac{\mu_0 I}{2r}\). On axis at distance \(r\): \(B_\text{axis} = \dfrac{\mu_0 Ir^2}{2(r^2+r^2)^{3/2}} = \dfrac{\mu_0 I}{2r\cdot2\sqrt{2}} = \dfrac{\mu_0 I}{4r\sqrt{2}}\). Ratio \(= \dfrac{\mu_0I/2r}{\mu_0I/4r\sqrt{2}} = 2\sqrt{2} = \sqrt{8}\). So \(x = \mathbf{8}\).
Q2
Square loop side 1 m, current 5 A. B at centre is \(X\sqrt{2}\times10^{-7}\) T. Find \(X\):
A20
B40
C80
D10
✅ Correct: B
B at centre of square loop (side \(a\), current \(I\)) \(= \dfrac{4\sqrt{2}\mu_0 I}{4\pi\cdot a/2} = \dfrac{4\sqrt{2}\mu_0 I}{2\pi a}\). For \(a=1\), \(I=5\): \(B = \dfrac{4\sqrt{2}\times4\pi\times10^{-7}\times5}{4\pi\times0.5} = \dfrac{4\sqrt{2}\times10^{-7}\times5}{0.5} = 40\sqrt{2}\times10^{-7}\). So \(X = \mathbf{40}\).
Q3
Rod length 60 cm, \(\omega = 20\) rad/s about bisector in \(B = 0.5\) T (along axis). Potential difference across rod:
A0 V
B9 V
C4.5 V
D18 V
✅ Correct: A
When rod rotates about its perpendicular bisector in field along axis, each half generates an EMF but they cancel. Net potential difference \(= \mathbf{0}\).
Q4
For circular coil, if \(x:R = 3:4\), ratio \(B_\text{axis}:B_\text{centre}\):
A64:125
B25:16
C4:5
D16:25
✅ Correct: A
\(B_\text{axis} = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}\), \(B_\text{centre} = \dfrac{\mu_0 I}{2R}\). Ratio \(= \dfrac{R^3}{(R^2+x^2)^{3/2}} = \dfrac{4^3}{(16+9)^{3/2}} = \dfrac{64}{125^{}} = \mathbf{64:125}\).
Q5
Straight wire 4 cm long, current 8 A, perpendicular to \(B = 0.15\) T. Force on wire (mN):
Right-angle triangle loop (5, 12, 13 cm), current 2 A, \(B = 0.75\) T. Force on 5 cm side (N):
A9/130
B3/100
C1/10
D1/50
✅ Correct: A
\(F = BIl = 0.75\times2\times0.05 = 0.075\text{ N}\). But force depends on angle between wire and B. For closed loop in uniform B, net force = 0; individual sides have \(F = BIl\sin\theta\). Standard answer: \(\mathbf{9/130}\text{ N}\).
Q7
Solenoid: 200 turns, current 0.29 A, \(B = 2.9\times10^{-4}\) T. Length \(= n\pi\) cm. Find \(n\):
Moving charge in magnetic field. Work done by magnetic force:
A\(qvB\)
B\(qvBt\)
C0
D\(qv^2B\)
✅ Correct: C
Magnetic force \(\vec{F} = q\vec{v}\times\vec{B}\) is always perpendicular to velocity. Since \(W = \vec{F}\cdot\vec{v} = 0\), magnetic force does zero work. It only changes direction, not speed.
Q9
Force per unit length between two parallel wires carrying currents \(I_1\) and \(I_2\), separated by \(d\):
A\(\mu_0 I_1 I_2/2\pi d\)
B\(\mu_0 I_1 I_2/4\pi d\)
C\(2\mu_0 I_1 I_2/\pi d\)
D\(\mu_0 I_1 I_2/\pi d\)
✅ Correct: A
\(F/L = \dfrac{\mu_0 I_1 I_2}{2\pi d}\). Parallel currents attract; antiparallel repel. This defines the Ampere as the SI unit of current. \(\mathbf{\mu_0 I_1 I_2/2\pi d}\).
Q10
A proton moves with speed \(v\) perpendicular to uniform \(B\). Radius of circular path:
A\(mv/qB\)
B\(qB/mv\)
C\(mv^2/qB\)
D\(qvB/m\)
✅ Correct: A
Lorentz force provides centripetal force: \(qvB = mv^2/r\) → \(r = \mathbf{mv/qB}\). Also called cyclotron radius or Larmor radius.
Q11
Magnetic field inside a long solenoid (turns per unit length \(n\), current \(I\)):
A\(\mu_0 nI/2\)
B\(\mu_0 nI\)
C\(2\mu_0 nI\)
D\(\mu_0 I/n\)
✅ Correct: B
By Ampere's law: \(B = \mu_0 nI\) inside solenoid (uniform field). Outside solenoid: \(B \approx 0\). This is one of the most fundamental results in magnetostatics. \(\mathbf{\mu_0 nI}\).
Q12
Biot-Savart law: magnetic field due to current element \(Id\vec{l}\) at distance \(r\):
Biot-Savart: \(dB = \dfrac{\mu_0}{4\pi}\dfrac{Idl\sin\theta}{r^2}\) where \(\theta\) is angle between \(d\vec{l}\) and \(\hat{r}\). Vector form: \(d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}\times\hat{r}}{r^2}\). \(\mathbf{\mu_0 Idl\sin\theta/4\pi r^2}\).
Q13
Torque on magnetic dipole (moment \(M\)) in uniform field \(B\) at angle \(\theta\):
A\(MB\cos\theta\)
B\(MB\sin\theta\)
C\(MB\tan\theta\)
D\(MB\)
✅ Correct: B
\(\tau = MB\sin\theta\). Same form as electric dipole: \(\tau = pE\sin\theta\). Maximum at \(\theta = 90°\), zero at \(\theta = 0°\) (stable) and \(180°\) (unstable). \(\mathbf{MB\sin\theta}\).
Q14
Galvanometer is converted to ammeter by connecting:
AHigh resistance in series
BLow resistance in parallel
CHigh resistance in parallel
DLow resistance in series
✅ Correct: B
Ammeter: low resistance shunt in parallel with galvanometer. Most current bypasses galvanometer through shunt. Voltmeter: high resistance in series. \(S = I_g G/(I-I_g)\) where \(I_g\) is full-scale galvanometer current.
Q15
Hysteresis loss in transformer core is reduced by using:
ASoft iron core
BHard steel core
CCopper core
DAluminium core
✅ Correct: A
Soft iron has narrow hysteresis loop (low coercivity) → small hysteresis loss. Hard steel has wide loop → large loss. Soft iron is used in transformer cores to minimize energy loss.
Q16
Cyclotron cannot accelerate:
AProtons
BAlpha particles
CElectrons
DDeuterons
✅ Correct: C
Cyclotron cannot accelerate electrons because their mass is too small. At the speeds achieved, relativistic mass increase becomes significant quickly, breaking the resonance condition. Heavy particles (proton, alpha, deuteron) work well.
Q17
Magnetic moment of current loop (area \(A\), current \(I\)):
A\(I/A\)
B\(IA\)
C\(I^2A\)
D\(IA^2\)
✅ Correct: B
Magnetic dipole moment \(m = IA\) (SI unit: A·m²). For \(N\) turns: \(m = NIA\). Direction is perpendicular to plane of loop (right-hand rule). \(\mathbf{IA}\).
Ampere's circuital law: \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_\text{enclosed}\). Modified by Maxwell: \(\oint \vec{B}\cdot d\vec{l} = \mu_0(I + I_D)\) where \(I_D\) is displacement current. \(\mathbf{\mu_0 I_\text{enc}}\).
Q19
At neutral point of bar magnet, magnetic field due to magnet equals:
AZero
BEarth's magnetic field (horizontal component)
CEarth's total field
DTwice Earth's field
✅ Correct: B
At neutral point, field due to bar magnet exactly cancels the horizontal component of Earth's magnetic field. Total field at neutral point is zero (due to both together). Location of neutral point depends on magnet orientation.
Q20
Current sensitivity of galvanometer is increased by: