Physics

Magnetism and Magnetic Effects

230 Questions

Magnetism and magnetic effects focus on the forces exerted by magnetic fields on moving charges and magnetic materials. Questions cover magnetic dipoles, flux density, the motion of charged particles, and electromagnetic relationships. It is a vital physics topic for government competitive exams.

Magnetic dipolesCharged particle motionMagnetic flux densityBar magnetsEarth magnetism

Magnetism and Magnetic Effects Questions

Multiple choice

A charge particle projected towards east is deflected towards north by a magnetic field. What is the direction of the magnetic field?

  1. Towards east

  2. Towards west

  3. Downwards

  4. Upwards

  5. -
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is the correct answer. According to Fleming's left hand rule: Hold out your left hand with forefinger, middle finger and thumb at right angle to one another. If the fore finger represents the direction of the field and the middle finger represents the direction of the current, then thumb gives the direction of the force. In this case, the direction of the charge particle is towards east, i.e. the direction of the current repesented by the fore finger and the thumb repesents the direction of the force. The fore finger represents the direction of the magnetic field.

Multiple choice

An electron moves in X-Y plane with a speed of 10^6 m/s. Its velocity vector makes an angle of 60 degrees with the X axis. A magnetic field of magnitude 10-2 T exists along Y axis. Compute the magnitude force exerted on the electron and its direction.

  1. 9 x 109 Newton along Z axis

  2. 8 x 10-16 Newton along Z axis

  3. 3.204 x 10-15 Newton along Y axis

  4. 8 x 103 Newton along Y axis

  5. -
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Flux Density (B) = 10-2 Wb Charge on electron (e) = 1.602 x 10-19 C

Multiple choice

A Hall effect element used for measuring magnetic field strength gives an output voltage of 10.5 mV. The element is made of silicon and is 2.5 mm thick and carries a current of a 4 A. The Hall coefficient is 4.1 x 10-6 Vm/A-wb/m2. Calculate the magnetic field strength.

  1. 1.6 Wb/m2

  2. 2.6 Wb/m2

  3. 0.63 Wb/m2

  4. 4.7 Wb/m2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hall effect element thickness , t = 2.5 mm = 2.5 x 10-3 m Output voltage(V H) = 10.5 m V = 10.5 x 10-3 V Current ( I) = 4A Hall coefficient (k H) = 4.1 x 10-6 Vm / A-wb/m2 Therefore, V H = (kH IB/) t B (magnetic field strength) = (V H x t)/(kH I) = (10.5 x 10-3 x 2.5 x 10-3)/(4.1 x 10-6 x 4) = 1.6 Wb/m2 

Multiple choice

The magnetic bearing of a line AB is S 45o E and the declination is 5o West. The true bearing of the line AB is

  1. S 45o E

  2. S 40o E

  3. S 50o E

  4. S 50o W
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Magnetic bearing is S 45 E. Declination is 5 degrees West, meaning the magnetic north is 5 degrees west of true north. To convert magnetic bearing to true bearing, subtract the westerly declination from the bearing angle relative to the north-south line. S 45 E becomes S (45 + 5) E = S 50 E.

Multiple choice properties of magnet magnetism physics neutral points magnetic poles and magnetic compass magnetic field

A DMM is arragned at the magnetic pole of earth in $\tan{A}$ position. If a bar magnet is placed at some distance from the needle, deflection is

  1. ${0}^{o}$
  2. ${90}^{o}$
  3. ${45}^{o}$
  4. ${180}^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At the magnetic poles, the horizontal component of the Earth's magnetic field (BH) is zero. In a tan A position, the needle aligns with the resultant field. Since BH = 0, the needle will be vertical (dip = 90 degrees), but the question asks for deflection in a horizontal DMM setup, which results in 0 degrees as there is no horizontal field to deflect it.

Multiple choice properties of magnet magnetism physics neutral points magnetic poles and magnetic compass magnetic field

If earth's magnetic field $B _H$, if the frequency of oscillation of a magnetic needle is n, then -

  1. $ n \propto B _H$
  2. $ n^2 \propto B _H$
  3. $ n \propto B _H^2$
  4. $n^2 \propto \frac{1}{B _H}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a magnetic needle oscillating in a magnetic field is T = 2 * pi * sqrt(I / (M * BH)). Since frequency n = 1/T, we have n = (1 / (2 * pi)) * sqrt(M * BH / I). Squaring both sides gives n^2 proportional to BH.

Multiple choice properties of magnet magnetism physics neutral points magnetic poles and magnetic compass magnetic field

In an uniform field the magnetic needle completes 10 oscillations in 92seconds. When a small magnet is placed in the magnetic meridian 10cm due north of needle with north pole towards south completes 15 oscillations in 69seconds. The magnetic moment of magnet ($B _H =0.3 \ G$) is

  1. $4.5 \ A m^2$
  2. $0.45 \ A m^2$
  3. $0.75 \ A m^2$
  4. $0.225 \ A m^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the oscillation magnetometer formula, the ratio of frequencies squared is proportional to the ratio of fields. T1 = 9.2s, T2 = 4.6s. (n2/n1)^2 = (T1/T2)^2 = (9.2/4.6)^2 = 4. The field B_net = BH + B_magnet. Solving for M using the tan A position formula yields 4.5 Am^2.

Multiple choice properties of magnet magnetism physics neutral points magnetic poles and magnetic compass magnetic field

A magnetic needle oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^0$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be 

  1. $\frac{T}{\sqrt 2}$

  2. T

  3. 2T

  4. 6T
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a horizontal plane, the period is T = 2 * pi * sqrt(I / (M * BH)). In a vertical plane, the effective field is the total magnetic field B = BH / cos(delta), where delta is the angle of dip. Thus, T_v = 2 * pi * sqrt(I / (M * B)) = T * sqrt(cos(delta)). With delta = 60 degrees, cos(60) = 1/2, so T_v = T / sqrt(2).

Multiple choice motional emf physics

The magnetic induction due to a magnet on the equatorial line at a distance 0.2 m is $54 \times 10^{-6}$T. The magnetic induction at 0.3m is

  1. $1.6 \times 10^{-6} \quad T$
  2. $1.6 \times 10^{-5} \quad T$
  3. $3.2 \times 10^{-6} \quad T$
  4. $3.2 \times 10^{-5} \quad T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnetic induction on the equatorial line is proportional to 1/r^3. B1 / B2 = (r2 / r1)^3. (54 * 10^-6) / B2 = (0.3 / 0.2)^3 = (1.5)^3 = 3.375. B2 = (54 * 10^-6) / 3.375 = 16 * 10^-6 = 1.6 * 10^-5 T.

Multiple choice motional emf physics

In the figure magnetic points into the plane of paper and the rod of length $l$ is moving in the field such that the bottom most point has a velocity $v _1$ and the topmost point has the velocity $V _2(V _2>V _1)$ The emf induced is given by 

  1. $Bv _1l$
  2. $Bv _2l$
  3. $\cfrac 1 2 B(v _2+v _1)l$
  4. $\cfrac 1 2 B(v _2-v _1)l$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Velocity of a point at a distance $x$ from the bottom is given by,

$v=v _1+(\dfrac{v _1-v _2}{l})x$

Potential difference on a small length at this distance is $e=Bx(dx)$

Therefore,

Total potential difference, $e=\int^l _0 Bvdx$

$e=\int^l _0 B(v _1+\dfrac{(v _2-v _1)x}{l})dx$

$=\dfrac{B(v _1+v _2)l}{2}$
Multiple choice motional emf physics

A uniform magnetic field exists in region given by $\vec B = 3\hat i + 4\hat j + 5\hat k$. A rod of length $5m$ is placed along $y$ moved along $x-axis$ with constant speed $1m/sec$. Then induced e.m.f. in the rod will be:

  1. $zero$
  2. $25V$
  3. $20V$
  4. $15V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The induced emf is given by E = (v x B) dot L. Here, v = 1 * i (m/s), B = 3i + 4j + 5k (T), and L = 5 * j (m). The cross product v x B = (1 * i) x (3i + 4j + 5k) = 4k - 5j. Then E = (4k - 5j) dot (5j) = -25V. The magnitude is 25V.

Multiple choice motional emf physics

The amplitude of a magnetic field, which is part of a harmonic electromagnetic wave in vacuum, is  $\mathrm { B } _ { 0 } =510\mathrm { nT } .$  What is the amplitude of the electric field of the wave? 

  1. $140 \mathrm { NC } ^ { - 1 }$
  2. $153 \mathrm { NC } ^ { - 1 }$
  3. $163 \mathrm { NC } ^ { - 1 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an electromagnetic wave, the relationship between electric field amplitude E_0 and magnetic field amplitude B_0 is E_0 = c * B_0, where c is the speed of light (3 * 10^8 m/s). E_0 = (3 * 10^8) * (510 * 10^-9) = 153 N/C.

Multiple choice motional emf physics

A charged particle enters in a uniform magnetic field with velocity at an angle of ${ 60 }^{ o }$ with the magnetic field. The pitch of helical path is x, the radius of helix is 

  1. $\dfrac { x }{ 2\sqrt { 3 } x } $
  2. $\dfrac { 2x }{ \sqrt { 3 } \pi } $
  3. $\dfrac { \sqrt { 3 } x }{ \pi } $
  4. $\dfrac { \sqrt { 3 } x }{ 2\pi } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pitch x = v * cos(theta) * T = v * cos(theta) * (2 * pi * m) / (q * B). Radius R = (m * v * sin(theta)) / (q * B). Dividing R by x gives R / x = (sin(theta)) / (2 * pi * cos(theta)) = tan(60) / (2 * pi) = sqrt(3) / (2 * pi). Thus R = (sqrt(3) * x) / (2 * pi).