Magnetic Field-Solved Numericals

1.A proton moving with a speed of 3.4*10^7m/s enters a mag. field in direction perpendicular to the field. The strength of magnetic field is 2T Calculate the force acting on the proton and the acceleration produced on it.

Given:

Velocity of a proton (v) = 3.4 * 107 m/s.

Magnetic field (B) = 2T

Angle between field and direction of beam θ = 90o

Charge on proton (a) = 1.6 * 10-19C.

F = BqVsinθ

= 2 * 1.6 * 10-19 * 3.4 * 107

= 1.67 * 10-11 N

Also, Acceleration (A) = Fm$\frac{\mathrm{F}}{\mathrm{m}}$

= 1.09∗10−111.67∗10−37$\frac{1.09\ast {10}^{-11}}{1.67\ast {10}^{-37}}$

= 6.5 8 1015m/s2.

2.An electron is moving vertically upward with a speed of 2*10^8m/s. What will be the magnitude of the force on the electron exerted by a horizontal magnetic field of 0.5T directed towards east?

Given:

Velocity of electron (v) = 2.0 * 108 m/s.

B = 0.5T

θ = 90o

F = BqVsinθ

= 0.5 * 1.6 * 10-16 * 2 * 108

= 1.6 * 10-11N.

3.A horizontal wire 0.1m long carries a current of 5A. Find the mag. field which can supports the weight of wire assuming that it's mass is 3*10^-3kg/m.

Given:

Length of wire (θ) = 0.1m

Current (I) = 5A

Mass of wire (m) = 3 * 10-3 kg/m

= 30 * 10-3 kg

We have,

F = BILsinθ

Or, ma = BIL

Or, B = mgIL$\frac{\mathrm{m}\mathrm{g}}{\mathrm{I}\mathrm{L}}$

= 30 * 10-3 *9.8/ 5*0.1 = 5.88*10-10T

4.What is the force of wire of length 4cm placed inside a solenoid near its centre making an angle of 60deg. with the axis? The wire carries current of 12A and the mag. field due to solenoid has a magnitude of 0.25T.

Given:

Length of wire (l) = 4cm

Angle (θ) = 60o+

Current (I) = 12A

B = 0.25T

Force along the axis of solenoid.

5.A horizontal overhead power line carries a current of 9A in east to west direction. What is the magnitude and direction of mag. field due to current 1.5m below the wire?

Soln:

Given:

Current (I) = 90A in east west direction.

Here, Magnetic field due to straight current carrying conductor.

Or, B = uoI2πa$\frac{{\mathrm{u}}_{\mathrm{o}}\mathrm{I}}{2\pi \mathrm{a}}$

= 1.25∗10−6∗902π∗1.5$\frac{1.25\ast {10}^{-6}\ast 90}{2\pi \ast 1.5}$

= 1.2 8 10-5 T towards south.

6.A circular segment of radius 10cm subtends an angle of 60deg and it's centre. A current of 9A is flowing through it. Find the magnitude of magnetic also it;s direction produce at the centre.

Soln:

Given:

Radius (r) = 10cm

Angle (θ) = 60o

Current (I) = 9A

To find magnitude and direction of magnetic field produced at the center,

B = uoI2r∗θ2π$\frac{{\mathrm{u}}_{\mathrm{o}}\mathrm{I}}{2\mathrm{r}}\ast \frac{\theta }{2\pi }$

= 1.25∗10−6∗92∗10∗602∗360$\frac{1.25\ast {10}^{-6}\ast 9}{2\ast 10}\ast \frac{60}{2\ast 360}$

= 9.4 * 10-6T. Perpendicular to the plane of paper pointing downward.

7.A 10cm diameter circular loop of wire placed with it's face parallel to uniform mag. field between the poles of a large magnet. When 8.1A flows in the loop, the torque on it is 0.116Nm calculate the torque acting on the coil.

Soln:

Given:

Diameter (d) = 10cm

Current (I) = 8.1A

Torque (T) = 0.116Nm

We know,

Torque due to magnetic field,

T = BINAcosθ

Or, 0.116 = B * 8.1 * 1 * π (0.12)2${\left(\frac{0.1}{2}\right)}^2$ * cos0o

Or, B = 0.1168.1∗1∗0.0025$\frac{0.116}{8.1\ast 1\ast 0.0025}$

So, B = 1.8T.

8.The plane of a rectangular coil makes an angle of 60deg with a direction of a uniform mag.field between the poles of 4*10^-2T. The coil is of 20turns, measuring 20cm by 100cm and carries a current of 0.5A, calculate the torque acting on the coil.

Soln:

Given:

Angle with magnetic filed (θ) = 60o.

Strength (B) = 4 8 10-2T

No. of turns (N) = 20

Area of coil = 0.2 * 1

= 0.2 m2

Current (I) = 0.5A

Now, Torque on a coil (T) = BINA cosθ

= 4 * 10-2 * 0.5 * 20 * 0.2 cosθ

= 4 * 10-3Nm.

9.A 32cm long solenoid,1.2cm is diameter is to produce a 0.2T mag. field at it's centre. If current in the solenoid is 3.7A, how many turns must the solenoid have?

Soln:

 Given:

Length of solenoid (l) = 0.32m

Diameter (d) = 0.012m

Magnetic field (B) = 0.2T

Current (I) = 3.7A

No. of turns (N) = ?

We know,

B = uoNI /2π r

10.A toroid has a core of inner 25cm and other 26 cm around which 350deg turns of a wire are wound. If a current in the wire is 11A, calculate mag. field 

i) inside the toroid

ii) outside the toroid

Soln:

Given:

Inner radius (r1) = 0.25m

Outer radius (r2) = 0.26m

Angle (θ) = 350o

Current (I) = 11A

Now,

Magnetic field inside the toroid.

B = uoNI

= 4π * 10-7 * 1 * 11.


11. Please see the book
ans:

Here,

Mass of proton (mp) = 1.7 * 10-24 kg.

Radius of earth (r) = 6.37 * 106m.

Velocity(V) = 1.0 * 107 ms-1.

Now, BeV = mv2r

Or, B = mver = 1.7∗10−24∗1.0∗1071.6∗10−19∗6.37∗106

So, B = 1.67 * 10-8T.

(i) Length(l) = 1.5m, diameter = 0.4cm = 4 * 10-3m

No. of turns (n) = 10 turns/cm = 1000 turns/m.

Current (I) = 5.0A.

B = uonI = 4π * 10-7 * 1000 * 5.0

So, B = 6.28 * 10-3T.

(ii) radius of solenoid (r) = 4∗10−32 = 2 * 10-3m.

We know, B = u­onI

= 4π * 10-7 * 10001.5* 5

= 4.18 * 10-3T.

12. Please see the book.

ans:

Thickness of slab(d) =2mm=2 x 10-3m

Wide of slabe (b)=1.5 x 10-2m

Area of slab(A)=b x d =3 x 10-5m2

Current (I)=75A

p.d(V)=0.81μV=0.81 x 10-6V

concentration of electron per unit volume (n)=?

Now current density on x-axis Jx=I/A=753x10−5=25 x 105A/m2

Again magnetic field on the x- axis direction we have

Ex= V/d =0.81x10−61.5x10−2= 5.4 x 10-5V/m.

Now concentration of electron per unit volume is given by

n=−Jx.BxqEx=0.40∗2.5∗1061.6∗10−19∗5.4∗10−5= =1.16∗1028/m3

13.Please see the book.

ans:

Here,

Length (l) = 0.5m.

Horizontal magnetic field (B) = 0.1 T.

(i) Current (I) = 4A

Force on X (F) = BIL

= 0.1 * 4 * 0.5 = 0.2N.

(ii) Force (F) = BIL.sin(90 – θ)

Or, 0.22=0.1∗4∗0.5∗sin(90∘−θ)

Or, 0.1 = 0.2cosθ.

Or, cosθ = 0.10.2 = 12

Or, θ = cos-1(12)= 60°.

14.Please see the book

mass (m) = 50kg,

length of rod (l) = 0.5m

Magnetic flux(B) = 0.2T.

Current (I) = ?.

For the weight of the rod to be balanced by the force on the conductor due to the current flowing through it, BIL sinθ = mg.

Here, θ = 90°

BIL = mg

I = mgIL=0.050∗100.2∗0.5 = 5A.

15.

Number of turns (N) = 10

Case I: resistance of galvanometer (Rg) = 4 ohm.

Or, (θI)A=BNAC

Or, (θI)A=B∗10∗AC….(i)

Case II n = 100 turns.

Or, (θI)B=BNAC

Or, (θI)B=100BAC….(ii)

Performing (i)/(ii),

Or, (θI)A(θI)B=110

Hence, the current sensitivity changes by 10.

From (i), (θI)A=B∗10∗AC

Or, (θV)A=B∗10∗AC∗R

Or, (θV)A=10BA4C…..(iii),

From (ii),

Or, (θI)B=100BAC

Or, (θV)B=100BA160C….(iv)

Performing (iv)/(iii).

Or, (θV)B(θV)A=100BA160C10BA4C

= 100160∗410

= 14

So, (θV)B(θV)A=14

So, voltage sensitivity increases by 14.

If you find any mistakes then please let me know in the comment section below.

Thanks!!